CONTENTS
CONTENTS Chapter 1
Introduction to Statistics
Chapter 2
Descriptive Statistics
11
Chapter 3
Probability
71
Chapter 4
Discrete Probability Distributions
97
Chapter 5
Normal Probability Distributions
119
Chapter 6
Confidence Intervals
159
Chapter 7
Hypothesis Testing with One Sample
179
Chapter 8
Hypothesis Testing with Two Samples
215
Chapter 9
Correlation and Regression
253
Chapter 10
ChiSquare Tests and the FDistribution
287
Chapter 11
Nonparametric Tests
325
Appendix A
Alternative Presentation of the Standard Normal Distribution
351
Normal Probability Plots and Their Graphs
352
Appendix C
1
Activities
353
Case Studies
357
Uses and Abuses
367
Real Statistics–Real Decisions
371
Technologies
381
Introduction to Statistics
CHAPTER
1
1.1 AN OVERVIEW OF STATISTICS
1.1 Try It Yourself Solutions 1a. The population consists of the prices per gallon of regular gasoline at all gasoline stations in the United States. b. The sample consists of the prices per gallon of regular gasoline at the 800 surveyed stations. c. The data set consists of the 800 prices. 2a. Because the numerical measure of $2,326,706,685 is based on the entire collection of player’s salaries, it is from a population. b. Because the numerical measure is a characteristic of a population, it is a parameter. 3a. Descriptive statistics involve the statement “76% of women and 60% of men had a physical examination within the previous year.” b. An inference drawn from the study is that a higher percentage of women had a physical examination within the previous year.
1.1 EXERCISE SOLUTIONS 1. A sample is a subset of a population. 2. It is usually impractical (too expensive and time consuming) to obtain all the population data. 3. A parameter is a numerical description of a population characteristic. A statistic is a numerical description of a sample characteristic. 4. Descriptive statistics and inferential statistics. 5. False. A statistic is a numerical measure that describes a sample characteristic. 6. True 7. True 8. False. Inferential statistics involves using a sample to draw conclusions about a population. 9. False. A population is the collection of all outcomes, responses, measurements, or counts that are of interest. 10. True 11. The data set is a population because it is a collection of the ages of all the members of the House of Representatives. 12. The data set is a sample because only every fourth person is measured. 13. The data set is a sample because the collection of the 500 spectators is a subset within the population of the stadium’s 42,000 spectators. 14. The data set is a population because it is a collection of the annual salaries of all lawyers at a firm.
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CHAPTER 1

INTRODUCTION TO STATISTICS
15. Sample, because the collection of the 20 patients is a subset within the population. 16. The data set is a population since it is a collection of the number of televisions in all U.S. households. 17. Population: Party of registered voters in Warren County. Sample: Party of Warren County voters responding to phone survey. 18. Population: Major of college students at Caldwell College. Sample: Major of college students at Caldwell College who take statistics. 19. Population: Ages of adults in the United States who own computers. Sample: Ages of adults in the United States who own Dell computers. 20. Population: Income of all homeowners in Texas. Sample: Income of homeowners in Texas with mortgages. 21. Population: All adults in the United States that take vacations. Sample: Collection of 1000 adults surveyed that take vacations. 22. Population: Collection of all infants in Italy. Sample: Collection of the 33,043 infants in the study. 23. Population: Collection of all households in the U.S. Sample: Collection of 1906 households surveyed. 24. Population: Collection of all computer users. Sample: Collection of 1000 computer users surveyed. 25. Population: Collection of all registered voters. Sample: Collection of 1045 registered voters surveyed. 26. Population: Collection of all students at a college. Sample: Collection of 496 college students surveyed. 27. Population: Collection of all women in the U.S. Sample: Collection of the 546 U.S. women surveyed. 28. Population: Collection of all U.S. vacationers. Sample: Collection of the 791 U.S. vacationers surveyed. 29. Statistic. The value $68,000 is a numerical description of a sample of annual salaries. 30. Statistic. 43% is a numerical description of a sample of high school students. 31. Parameter. The 62 surviving passengers out of 97 total passengers is a numerical description of all of the passengers of the Hindenburg that survived. 32. Parameter. 44% is a numerical description of the total number of governors. 33. Statistic. 8% is a numerical description of a sample of computer users. 34. Parameter. 12% is a numerical description of all new magazines.
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CHAPTER 1

INTRODUCTION TO STATISTICS
35. Statistic. 53% is a numerical description of a sample of people in the United States. 36. Parameter. 21.1 is a numerical description of ACT scores for all graduates. 37. The statement “56% are the primary investor in their household” is an application of descriptive statistics. An inference drawn from the sample is that an association exists between U.S. women and being the primary investor in their household. 38. The statement “spending at least $2000 for their next vacation” is an application of descriptive statistics. An inference drawn from the sample is that U.S. vacationers are associated with spending more than $2000 for their next vacation. 39. Answers will vary. 40. (a) The volunteers in the study represent the sample. (b) The population is the collection of all individuals who completed the math test. (c) The statement “three times more likely to answer correctly” is an application of descriptive statistics. (d) An inference drawn from the sample is that individuals who are not sleep deprived will be three times more likely to answer math questions correctly than individuals who are sleep deprived. 41. (a) An inference drawn from the sample is that senior citizens who live in Florida have better memory than senior citizens who do not live in Florida. (b) It implies that if you live in Florida, you will have better memory. 42. (a) An inference drawn from the sample is that the obesity rate among boys ages 2 to 19 is increasing. (b) It implies the same trend will continue in future years. 43. Answers will vary.
1.2 DATA CLASSIFICATION
1.2 Try It Yourself Solutions 1a. One data set contains names of cities and the other contains city populations. b. City: Nonnumerical Population: Numerical c. City: Qualitative Population: Quantitative
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2a. (1) The final standings represent a ranking of basketball teams. (2) The collection of phone numbers represents labels. No mathematical computations can be made. b. (1) Ordinal, because the data can be put in order. (2) Nominal, because you cannot make calculations on the data. 3a. (1) The data set is the collection of body temperatures. (2) The data set is the collection of heart rates. b. (1) Interval, because the data can be ordered and meaningful differences can be calculated, but it does not make sense writing a ratio using the temperatures. (2) Ratio, because the data can be ordered, can be written as a ratio, you can calculate meaningful differences, and the data set contains an inherent zero.
1.2 EXERCISE SOLUTIONS 1. Nominal and ordinal
2. Ordinal, Interval, and Ratio
3. False. Data at the ordinal level can be qualitative or quantitative. 4. False. For data at the interval level, you can calculate meaningful differences between data entries. You cannot calculate meaningful differences at the nominal or ordinal level. 5. False. More types of calculations can be performed with data at the interval level than with data at the nominal level. 6. False. Data at the ratio level can be placed in a meaningful order. 7. Qualitative, because telephone numbers are merely labels. 8. Quantitative, because the daily high temperature is a numerical measure. 9. Quantitative, because the lengths of songs on an MP3 player are numerical measures. 10. Qualitative, because the player numbers are merely labels. 11. Qualitative, because the poll results are merely responses. 12. Quantitative, because the diastolic blood pressure is a numerical measure. 13. Qualitative. Ordinal. Data can be arranged in order, but differences between data entries make no sense. 14. Qualitative. Nominal. No mathematical computations can be made and data are categorized using names. 15. Qualitative. Nominal. No mathematical computations can be made and data are categorized using names. 16. Quantitative. Ratio. A ratio of two data values can be formed so one data value can be expressed as a multiple of another. 17. Qualitative. Ordinal. The data can be arranged in order, but differences between data entries are not meaningful. 18. Quantitative. Ratio. The ratio of two data values can be formed so one data value can be expressed as a multiple of another.
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CHAPTER 1
19. Ordinal
20. Ratio
21. Nominal
22. Ratio
23. (a) Interval
(b) Nominal
(c) Ratio
(d) Ordinal
24. (a) Interval
(b) Nominal
(c) Interval
(d) Ratio

INTRODUCTION TO STATISTICS
25. An inherent zero is a zero that implies “none.” Answers will vary. 26. Answers will vary.
1.3 EXPERIMENTAL DESIGN
1.3 Try It Yourself Solutions 1a. (1) Focus: Effect of exercise on relieving depression. (2) Focus: Success of graduates. b. (1) Population: Collection of all people with depression. (2) Population: Collection of all university graduates. c. (1) Experiment (2) Survey 2a. There is no way to tell why people quit smoking. They could have quit smoking either from the gum or from watching the DVD. b. Two experiments could be done; one using the gum and the other using the DVD. 3a. Example: start with the first digits 92630782 . . .
ⱍ ⱍ ⱍ ⱍ ⱍ ⱍ
b. 92 63 07 82 40 19 26 c. 63, 7, 40, 19, 26 4a. (1) The sample was selected by only using available students. (2) The sample was selected by numbering each student in the school, randomly choosing a starting number, and selecting students at regular intervals from the starting number. b. (1) Because the students were readily available in your class, this is convenience sampling. (2) Because the students were ordered in a manner such that every 25th student is selected, this is systematic sampling.
1.3 EXERCISE SOLUTIONS 1. In an experiment, a treatment is applied to part of a population and responses are observed. In an observational study, a researcher measures characteristics of interest of part of a population but does not change existing conditions. 2. A census includes the entire population; a sample includes only a portion of the population. 3. Assign numbers to each member of the population and use a random number table or use a random number generator.
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4. Replication is the repetition of an experiment using a large group of subjects. It is important because it gives validity to the results. 5. True 6. False. A doubleblind experiment is used to decrease the placebo effect. 7. False. Using stratified sampling guarantees that members of each group within a population will be sampled. 8. False. A census is a count of an entire population. 9. False. To select a systematic sample, a population is ordered in some way and then members of the population are selected at regular intervals. 10. True 11. In this study, you want to measure the effect of a treatment (using a fat substitute) on the human digestive system. So, you would want to perform an experiment. 12. It would be nearly impossible to ask every consumer whether he or she would still buy a product with a warning label. So, you should use a survey to collect these data. 13. Because it is impractical to create this situation, you would want to use a simulation. 14. Because the U.S. Congress keeps accurate financial records of all members, you could take a census. 15. (a) The experimental units are the 30–35 year old females being given the treatment. (b) One treatment is used. (c) A problem with the design is that there may be some bias on the part of the researchers if he or she knows which patients were given the real drug. A way to eliminate this problem would be to make the study into a doubleblind experiment. (d) The study would be a doubleblind study if the researcher did not know which patients received the real drug or the placebo. 16. (a) The experimental units are the people with early signs of arthritis. (b) One treatment is used. (c) A problem with the design is that the sample size is small. The experiment could be replicated to increase validity. (d) In a placebocontrolled doubleblind experiment, neither the subject nor the experimenter knows whether the subject is receiving a treatment or a placebo. The experimenter is informed after all the data have been collected. (e) The group could be randomly split into 20 males or 20 females in each treatment group. 17. Each U.S. telephone number has an equal chance of being dialed and all samples of 1599 phone numbers have an equal chance of being selected, so this is a simple random sample. Telephone sampling only samples those individuals who have telephones, are available, and are willing to respond, so this is a possible source of bias. 18. Because the persons are divided into strata (rural and urban), and a sample is selected from each stratum, this is a stratified sample.
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CHAPTER 1

INTRODUCTION TO STATISTICS
19. Because the students were chosen due to their convenience of location (leaving the library), this is a convenience sample. Bias may enter into the sample because the students sampled may not be representative of the population of students. For example, there may be an association between time spent at the library and drinking habits. 20. Because the disaster area was divided into grids and thirty grids were then entirely selected, this is a cluster sample. Certain grids may have been much more severely damaged than others, so this is a possible source of bias. 21. Because a random sample of outpatients were selected and all samples of 1210 patients had an equal chance of being selected, this is a simple random sample. 22. Because every twentieth engine part is sampled from an assembly line, this is a systematic sample. It is possible for bias to enter into the sample if, for some reason, the assembly line performs differently on a consistent basis. 23. Because a sample is taken from each oneacre subplot (stratum), this is a stratified sample. 24. Because a sample is taken from members of a population that are readily available, this is a convenience sample. The sample may be biased if the teachers sampled are not representative of the population of teachers. For example, some teachers may frequent the lounge more often than others. 25. Because every ninth name on a list is being selected, this is a systematic sample. 26. Each telephone has an equal chance of being dialed and all samples of 1012 phone numbers have an equal chance of being selected, so this is a simple random sample. Telephone sampling only samples those individuals who have telephones, are available, and are willing to respond, so this is a possible source of bias. 27. Answers will vary. 28. Answers will vary. 29. Census, because it is relatively easy to obtain the salaries of the 50 employees. 30. Sampling, because the population of students is too large to easily record their color. Random sampling would be advised since it would be easy to randomly select students then record their favorite car color. 31. Question is biased because it already suggests that drinking fruit juice is good for you. The question might be rewritten as “How does drinking fruit juice affect your health?” 32. Question is biased because it already suggests that drivers who change lanes several times are dangerous. The question might be rewritten as “Are drivers who change lanes several times dangerous?” 33. Question is unbiased because it does not imply how many hours of sleep are good or bad. 34. Question is biased because it already suggests that the media has a negative effect on teen girls’ dieting habits. The question might be rewritten as “Do you think the media has an effect on teen girls’ dieting habits?” 35. The households sampled represent various locations, ethnic groups, and income brackets. Each of these variables is considered a stratum. 36. Stratified sampling ensures that each segment of the population is represented.
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INTRODUCTION TO STATISTICS
37. Open Question Advantage: Allows respondent to express some depth and shades of meaning in the answer. Disadvantage: Not easily quantified and difficult to compare surveys. Closed Question Advantage: Easy to analyze results. Disadvantage: May not provide appropriate alternatives and may influence the opinion of the respondent. 38. (a) Advantage: Usually results in a savings in the survey cost. (b) Disadvantage: There tends to be a lower response rate and this can introduce a bias into the sample. Sampling Technique: Convenience sampling 39. Answers will vary. 40. If blinding is not used, then the placebo effect is more likely to occur. 41. The Hawthorne effect occurs when a subject changes behavior because he or she is in an experiment. However, the placebo effect occurs when a subject reacts favorably to a placebo he or she has been given. 42. Both a randomized block design and a stratified sample split their members into groups based on similar characteristics. 43. Answers will vary.
CHAPTER 1 REVIEW EXERCISE SOLUTIONS 1. Population: Collection of all U.S. adults. Sample: Collection of the 1000 U.S. adults that were sampled. 2. Population: Collection of all nurses in San Francisco area. Sample: Collection of 38 nurses in San Francisco area that were sampled. 3. Population: Collection of all credit cards. Sample: Collection of 146 credit cards that were sampled. 4. Population: Collection of all physicians in the U.S. Sample: Collection of 1205 physicians that were sampled. 5. The team payroll is a parameter since it is a numerical description of a population (entire baseball team) characteristic. 6. Since 42% is describing a characteristic of the sample, this is a statistic. 7. Since “10 students” is describing a characteristic of a population of math majors, it is a parameter. 8. Since 19% is describing a characteristic of a sample of Indiana ninth graders, this is a statistic. 9. The average late fee of $27.46 charged by credit cards is representative of the descriptive branch of statistics. An inference drawn from the sample is that all credit cards charge a late fee of $27.46.
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CHAPTER 1

INTRODUCTION TO STATISTICS
10. 60% of all physicians surveyed consider leaving the practice of medicine because they are discouraged over the state of U.S. healthcare is representative of the descriptive branch of statistics. An inference drawn from the sample is that 60% of all physicians surveyed consider leaving the practice of medicine because they are discouraged over the state of U.S. healthcare. 11. Quantitative because monthly salaries are numerical measurements. 12. Qualitative because Social Security numbers are merely labels for employees. 13. Quantitative because ages are numerical measurements. 14. Qualitative because zip codes are merely labels for the customers. 15. Interval. It makes no sense saying that 100 degrees is twice as hot as 50 degrees. 16. Ordinal. The data are qualitative but could be arranged in order of car size. 17. Nominal. The data are qualitative and cannot be arranged in a meaningful order. 18. Ratio. The data are numerical, and it makes sense saying that one player is twice as tall as another player. 19. Because CEOs keep accurate records of charitable donations, you could take a census. 20. Because it is impractical to create this situation, you would want to perform a simulation. 21. In this study, you want to measure the effect of a treatment (fertilizer) on a soybean crop. You would want to perform an experiment. 22. Because it would be nearly impossible to ask every college student about his/her opinion on environmental pollution, you should take a survey to collect the data. 23. The subjects could be split into male and female and then be randomly assigned to each of the five treatment groups. 24. Number the volunteers and then use a random number generator to randomly assign subjects to one of the treatment groups or the control group. 25. Because random telephone numbers were generated and called, this is a simple random sample. 26. Because the student sampled a convenient group of friends, this is a convenience sample. 27. Because each community is considered a cluster and every pregnant woman in a selected community is surveyed, this is a cluster sample. 28. Because every third car is stopped, this is a systematic sample. 29. Because grade levels are considered strata and 25 students are sampled from each stratum, this is a stratified sample. 30. Because of the convenience of surveying people waiting for their baggage, this is a convenience sample. 31. Telephone sampling only samples individuals who have telephones, are available, and are willing to respond. 32. Due to the convenience sample taken, the study may be biased toward the opinions of the student’s friends. 33. The selected communities may not be representative of the entire area.
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INTRODUCTION TO STATISTICS
32. It may be difficult for the law enforcement official to stop every third car.
CHAPTER 1 QUIZ SOLUTIONS 1. Population: Collection of all individuals with anxiety disorders. Sample: Collection of 372 patients in study. 2. (a) Statistic. 19% is a characteristic of a sample of Internet users. (b) Parameter. 84% is a characteristic of the entire company (population). (c) Statistic. 40% is a characteristic of a sample of Americans. 3. (a) Qualitative, since post office box numbers are merely labels. (b) Quantitative, since a final exam is a numerical measure. 4. (a) Nominal. Badge numbers may be ordered numerically, but there is no meaning in this order and no mathematical computations can be made. (b) Ratio. It makes sense to say that the number of candles sold during the 1st quarter was twice as many as sold in the 2nd quarter. (c) Interval because meaningful differences between entries can be calculated, but a zero entry is not an inherent zero. 5. (a) In this study, you want to measure the effect of a treatment (low dietary intake of vitamin C and iron) on lead levels in adults. You want to perform an experiment. (b) Because it would be difficult to survey every individual within 500 miles of your home, sampling should be used. 6. Randomized Block Design 7. (a) Because people were chosen due to their convenience of location (on the campground), this is a convenience sample. (b) Because every tenth part is selected from an assembly line, this is a systematic sample. (c) Stratified sample because the population is first stratified and then a sample is collected from each stratum. 8. Convenience
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CHAPTER
Descriptive Statistics
2
2.1 FREQUENCY DISTRIBUTIONS AND THEIR GRAPHS
2.1 Try It Yourself Solutions 1a. The number of classes (8) is stated in the problem. 89 15 b. Min 15 Max 89 Class width 9.25 ⇒ 10 8 c. Lower limit Upper limit 15 25 35 45 55 65 75 85
24 34 44 54 64 74 84 94
d. See part (e). e.
Class
Frequency, f
15–24 25–34 35–44 45–54 55–64 65–74 75–84 85–94
16 34 30 23 13 2 0 1
2a. See part (b). b. Class
Frequency, f
Midpoint
Relative frequency
Cumulative frequency
15–24 25–34 35–44 45–54 55–64 65–74 75–84 85–94
16 34 30 23 13 2 0 1
19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5
0.13 0.29 0.25 0.19 0.11 0.02 0.00 0.01
16 50 80 103 116 118 118 119
f 119
f
n1
c. 86% of the teams scored fewer than 55 touchdowns. 3% of the teams scored more than 65 touchdowns.
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CHAPTER 2
3a.

DESCRIPTIVE STATISTICS
Class Boundaries 14.5–24.5 24.5–34.5 34.5–44.5 44.5–54.5 54.5–64.5 64.5–74.5 74.5–84.5 84.5–94.5
c.
Touchdowns Scored Frequency
36
b. Use class midpoints for the horizontal scale and frequency for the vertical scale.
d. 86% of the teams scored fewer than 55 touchdowns. 3% of the teams scored more than 65 touchdowns.
30 24 18 12
19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5
6
Number of touchdowns
4a. Use class midpoints for the horizontal scale and frequency for the vertical scale. b. See part (c). c.
Touchdowns Scored Frequency
36 30 24 18 12
9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5
6
Number of touchdowns
d. The number of touchdowns increases until 34.5 touchdowns, then decreases afterward. 5abc. 0.30 0.25 0.20 0.15 0.10 0.05
19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5
Relative Frequency
Touchdowns Scored
Number of touchdowns
6a. Use upper class boundaries for the horizontal scale and cumulative frequency for the vertical scale. b. See part (c). c.
d. Approximately 80 teams scored 44 or fewer touchdowns.
100
e. Answers will vary.
80 60 40 20
14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5
Cumulative frequency
Touchdowns Scored 120
Number of touchdowns
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CHAPTER 2
7ab.
 DESCRIPTIVE STATISTICS
36
14.5
94.5 0
2.1 EXERCISE SOLUTIONS 1. By organizing the data into a frequency distribution, patterns within the data may become more evident. 2. Sometimes it is easier to identify patterns of a data set by looking at a graph of the frequency distribution. 3. Class limits determine which numbers can belong to that class. Class boundaries are the numbers that separate classes without forming gaps between them. 4. Cumulative frequency is the sum of the frequency for that class and all previous classes. Relative frequency is the proportion of entries in each class. 5. False. Class width is the difference between the lower and upper limits of consecutive classes. 6. True 7. False. An ogive is a graph that displays cumulative frequency. 8. True Max Min 58 7 8.5 ⇒ 9 Classes 6 Lower class limits: 7, 16, 25, 34, 43, 52
9. Width
Upper class limits: 15, 24, 33, 42, 51, 60 Max Min 94 11 10.375 ⇒ 11 Classes 8 Lower class limits: 11, 22, 33, 44, 55, 66, 77, 88
10. Width
Upper class limits: 21, 32, 43, 54, 65, 76, 87, 98 Max Min 123 15 18 ⇒ 19 Classes 6 Lower class limits: 15, 34, 53, 72, 91, 110
11. Width
Upper class limits: 33, 52, 71, 90, 109, 128 Max Min 171 24 14.7 ⇒ 15 Classes 10 Lower class limits: 24, 39, 54, 69, 84, 99, 114, 129, 144, 159
12. Width
Upper class limits: 38, 53, 68, 83, 98, 113, 128, 143, 158, 173
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CHAPTER 2

DESCRIPTIVE STATISTICS
13. (a) Class width ⫽ 31 ⫺ 20 ⫽ 11 (b) and (c) Class
Frequency, f
Midpoint
Class boundaries
20–30 31–41 42–52 53–63 64–74 75–85 86–96
19 43 68 69 74 68 24
25 36 47 58 69 80 91
19.5–30.5 30.5–41.5 41.5–52.5 52.5–63.5 63.5–74.5 74.5–85.5 85.5–96.5
f ⫽ 365 14a. Class width ⫽ 10 ⫺ 0 ⫽ 10 bc.
Class
Frequency
Midpoint
Class boundaries
0–9 10–19 20–29 30–39 40–49 50–59 60–69
188 372 264 205 83 76 32
4.5 14.5 24.5 34.5 44.5 54.5 64.5
⫺ 0.5–9.5 9.5–19.5 19.5–29.5 29.5–39.5 39.5–49.5 49.5–59.5 59.5–69.5
f ⫽ 1220
15. Class
Frequency, f
Midpoint
20–30 31–41 42–52 53–63 64–74 75–85 86–96
19 43 68 69 74 68 24
25 36 47 58 69 80 91
f ⫽ 365
Relative frequency
16.
0.05 0.12 0.19 0.19 0.20 0.19 0.07 f ⫽1 n
Cumulative frequency 19 62 130 199 273 341 365
Class
Frequency
Midpoint
Relative frequency
Cumulative frequency
0–9 10–19 20–29 30–39 40–49 50–59 60–69
188 372 264 205 83 76 32
4.5 14.5 24.5 34.5 44.5 54.5 64.5
0.15 0.30 0.22 0.17 0.07 0.06 0.03
188 560 824 1029 1112 1188 1220
f ⫽ 1220
17. (a) Number of classes ⫽ 7 (c) Greatest frequency 300 18. (a) Number of classes ⫽ 7 (c) Greatest frequency 900
f
n ⫽ 1 (b) Least frequency 10 (d) Class width ⫽ 10 (b) Least frequency 100 (d) Class width ⫽ 5
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CHAPTER 2
19. (a) 50
(b) 22.5–24.5 lbs
20. (a) 50
(b) 64–66 inches
21. (a) 24
(b) 29.5 lbs
22. (a) 44
(b) 66 inches
 DESCRIPTIVE STATISTICS
23. (a) Class with greatest relative frequency: 8–9 inches. Class with least relative frequency: 17–18 inches. (b) Greatest relative frequency 0.195 Least relative frequency 0.005 (c) Approximately 0.015 24. (a) Class with greatest relative frequency: 19–20 minutes. Class with least relative frequency: 21–22 minutes. (b) Greatest relative frequency 40% Least relative frequency 2% (c) Approximately 33% 25. Class with greatest frequency: 500–550 Class with least frequency: 250–300 and 700–750 26. Class with greatest frequency: 7.75–8.25 Class with least frequency: 6.25–6.75 Max Min 39 0 7.8 ⇒ 8 Number of classes 5
27. Class width
Class
Frequency, f
Midpoint
0–7 8–15 16–23 24–31 32–39
8 8 3 3 3
3.5 11.5 19.5 27.5 35.5
f 25
Relative frequency
Cumulative frequency
0.32 0.32 0.12 0.12 0.12 f 1 n
8 16 19 22 25
Class with greatest frequency: 0–7, 8–15 Class with least frequency: 16–23, 24–31, 32–39 28. Class width
Class 30–113 114–197 198–281 282–365 366–449 450–533
530 30 Max Min 83.3 ⇒ 84 Number of classes 6
Frequency
Midpoint
Relative frequency
Cumulative frequency
5 7 8 2 3 4
71.5 155.5 239.5 323.5 407.5 491.5
0.1724 0.2414 0.2759 0.0690 0.1034 0.1379
5 12 20 22 25 29
f 29
f 1 n
Class with greatest frequency: 198–281 Class with least frequency: 282–365
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15
CHAPTER 2

DESCRIPTIVE STATISTICS
29. Class width
Class
Max Min 7119 1000 1019.83 ⇒ 1020 Number of classes 6
Frequency, f
Midpoint
12 3 2 3 1 1
1509.5 2529.5 3549.5 4569.5 5589.5 6609.5
1000–2019 2020–3039 3040–4059 4060–5079 5080–6099 6100–7119
f 22
Relative frequency
Cumulative frequency
0.5455 0.1364 0.0909 0.1364 0.0455 0.0455 f 1 n
12 15 17 20 21 22
July Sales for Representatives Frequency
16
14 12 10 8 6 4 2 1509.5 3549.5 5589.5
Sales (in dollars)
Class with greatest frequency: 1000–2019 Class with least frequency: 5080–6099; 6100–7119 30. Class width
Max Min 51 32 3.8 ⇒ 4 Number of classes 5
Frequency
Midpoint
Relative frequency
Cumulative frequency
32–35 36–39 40–43 44–47 48–51
3 9 8 3 1
33.5 37.5 41.5 45.5 49.5
0.1250 0.3750 0.3333 0.1250 0.0417
3 12 20 23 24
f 24
Pungencies of Peppers
Frequency
Class
9 8 7 6 5 4 3 2 1 33.5 37.5 41.5 45.5 49.5
f 1 n
Pungencies (in 1000s of Scoville units)
Class with greatest frequency: 36–39
Frequency, f
Midpoint
5 4 3 5 6 4 1 2
304.5 332.5 360.5 388.5 416.5 444.5 472.5 500.5
f 30
Relative frequency
Cumulative frequency
0.1667 0.1333 0.1000 0.1667 0.2000 0.1333 0.0333 0.0667 f 1 n
5 9 12 17 23 27 28 30
Reaction Times for Females 6 4 2
304.5 332.5 360.5 388.5 416.5 444.5 472.5 500.5
Class 291–318 319–346 347–374 375–402 403–430 431–458 459–486 487–514
Max Min 514 291 27.875 ⇒ 28 Number of classes 8
Frequency
31. Class width
Reaction times (in milliseconds)
Class with greatest frequency: 403–430
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 DESCRIPTIVE STATISTICS
CHAPTER 2
2888 2456 Max Min 86.4 ⇒ 87 Number of classes 5
32. Class width
Class
Frequency
Midpoint
7 3 2 4 9
2499 2586 2673 2760 2847
0.28 0.12 0.08 0.16 0.36
2456–2542 2543–2629 2630–2716 2717–2803 2804–2890
f 25
Cumulative frequency 7 10 12 16 25
Frequency
Pressure at Fracture Time
Relative frequency
10 9 8 7 6 5 4 3 2 1 2499
2673
2847
Pressure (in pounds per square inch)
f 1 n
Class with greatest frequency: 2804–2890 Class with least frequency: 2630–2716 Max Min 264 146 23.6 ⇒ 24 Number of classes 5
0.2308 0.3462 0.1154 0.2308 0.0769 f 1 n
6 15 18 24 26
f 26
Bowling Scores 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05
253.5
157.5 181.5 205.5 229.5 253.5
Cumulative frequency
205.5
6 9 3 6 2
146–169 170–193 194–217 218–241 242–265
Relative frequency
229.5
Midpoint
181.5
Frequency, f
157.5
Class
Relative frequency
33. Class width
Scores
Class with greatest relative frequency: 170–193 Class with least relative frequency: 242–265 Max Min 80 10 14 ⇒ 15 Number of classes 5
Class
Frequency
Midpoint
Relative frequency
Cumulative frequency
10–24 25–39 40–54 55–69 70–84
11 9 6 2 4
17 32 47 62 77
0.3438 0.2813 0.1875 0.0625 0.1250
11 20 26 28 32
ATM Withdrawals Relative frequency
34. Class width
0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 17 32 47 62 77
Dollars
f
f 32
n1
Class with greatest relative frequency: 10–24 Class with least relative frequency: 55–69 Max Min 52 33 3.8 ⇒ 4 Number of classes 5
34.5 38.5 42.5 46.5 50.5
f 26
0.3077 0.2308 0.1923 0.0769 0.1923 f 1 n
8 14 19 21 26
Tomato Plant Heights 0.35 0.30 0.25 0.20 0.15 0.10 0.05
50.5
8 6 5 2 5
Cumulative frequency
46.5
33–36 37–40 41–44 45–48 49–52
Relative frequency
42.5
Midpoint
38.5
Frequency, f
34.5
Class
Relative frequency
35. Class width
Heights (in inches)
Class with greatest relative frequency: 33–36 Class with least relative frequency: 45–48 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17
CHAPTER 2

DESCRIPTIVE STATISTICS
36. Class width
16 7 Max Min 1.8 ⇒ 2 Number of classes 5
Class
Frequency
Midpoint
Relative frequency
Cumulative frequency
6–7 8–9 10–11 12–13 14–15
3 10 6 6 1
6.5 8.5 10.5 12.5 14.5
0.12 0.38 0.23 0.23 0.04
3 13 19 25 26
Years of Service Relative frequency
18
0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 6.5 8.5 10.5 12.514.5
Years
f
f 26
n1
Class with greatest relative frequency: 8–9 Class with least relative frequency: 14–15 Max Min 73 52 3.5 ⇒ 4 Number of classes 6
Class
Frequency, f
Relative frequency
Cumulative frequency
52–55 56–59 60–63 64–67 68–71 72–75
3 3 9 4 4 1
0.125 0.125 0.375 0.167 0.167 0.042
3 6 15 19 23 24
f 24
n1
Retirement Ages Cumulative frequency
37. Class width
25 20 15 10 5 51.5
f
59.5
67.5
75.5
Ages
Location of the greatest increase in frequency: 60–63 Max Min 57 16 6.83 ⇒ 7 Number of classes 6
f 20
20 15 10
57.5
50.5
5
43.5
2 5 13 18 18 20
36.5
0.10 0.15 0.40 0.25 0.00 0.10
29.5
2 3 8 5 0 2
Daily Saturated Fat Intake
22.5
Frequency, f
16–22 23–29 30–36 37–43 44–50 51–57
Cumulative frequency
15.5
Class
Relative frequency
Cumulative frequency
38. Class width
Daily saturated fat intake (in grams)
f 1 n
Location of the greatest increase in frequency: 30–36 Max Min 18 2 2.67 ⇒ 3 Number of classes 6
Class
Frequency, f
2–4 5–7 8–10 11–13 14–16 17–19
9 6 7 3 2 1
f 28
Relative frequency
Cumulative frequency
0.3214 0.2143 0.2500 0.1071 0.0714 0.0357 f 1 n
9 15 22 25 27 28
Gallons of Gasoline Purchased Cumulative frequency
39. Class width
30 25 20 15 10 5 1.5
7.5
13.5
19.5
Gasoline (in gallons)
Location of the greatest increase in frequency: 2–4 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2
Class
Frequency, f
Relative frequency
1–5 6–10 11–15 16–20 21–25 26–30
5 9 3 4 2 1
0.2083 0.3750 0.1250 0.1667 0.0833 0.0417
f 24
Cumulative frequency 5 14 17 21 23 24
Cumulative frequency
29 1 Max Min 4.67 ⇒ 5 Number of classes 6
40. Class width
 DESCRIPTIVE STATISTICS
Length of Cellular Phone Calls 30 25 20 15 10 5 0.5
f 1 n
10.5
20.5
30.5
Length of call (in minutes)
Location of the greatest increase in frequency: 6–10
Class
Max Min 98 47 10.2 ⇒ 11 Number of classes 5
Frequency, f
Midpoint
Relative frequency
Cumulative frequency
1 1 5 8 5
52 63 74 85 96
0.05 0.05 0.25 0.40 0.25
1 2 7 15 20
47–57 58–68 69–79 80–90 91–101
f 20
Exam Scores 10
Frequency
41. Class width
8 6 4 2
f 1 N
41 52 63 74 85 96 107
Scores
Class with greatest frequency: 80–90 Classes with least frequency: 47–57 and 58–68 Class
Frequency, f
Midpoint
Relative frequency
Cumulative frequency
0–2 3–5 6–8 9–11 12–14 15–17
16 17 7 1 0 1
1 4 7 10 13 16
0.3810 0.4048 0.1667 0.0238 0.0000 0.0238
16 33 40 41 41 42
f 42
Number of Children of First 42 Presidents 20
Frequency
42.
15 10 5 − 2 1 4 7 10 13 16 19
f 1 N
Number of children
Classes with greatest frequency: 0–2 Classes with least frequency: 15–17
61–66 67–72 73–78 79–84 85–90 91–96 97–102 103–108
Frequency, f
Midpoint
1 3 6 10 5 2 2 1
63.5 69.5 75.5 81.5 87.5 93.5 99.5 105.5
f 30
Relative frequency 0.0333 0.1000 0.2000 0.3333 0.1667 0.0667 0.0667 0.0333 f 1 N
Daily Withdrawals 0.35 0.30 0.25 0.20 0.15 0.10 0.05 63.5 69.5 75.5 81.5 87.5 93.5 99.5 105.5
Class
104 61 Max Min 5.375 ⇒ 6 Number of classes 8 Relative frequency
43. (a) Class width
Dollars (in hundreds)
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DESCRIPTIVE STATISTICS
(b) 16.7%, because the sum of the relative frequencies for the last three classes is 0.167. (c) $9600, because the sum of the relative frequencies for the last two classes is 0.10.
Frequency
Midpoint
Relative frequency
1 1 4 6 8 6 9 5 7 3
457.5 553.5 649.5 745.5 841.5 937.5 1033.5 1129.5 1225.5 1321.5
0.02 0.02 0.08 0.12 0.16 0.12 0.18 0.10 0.14 0.06
410–505 506–601 602–697 698–793 794–889 890–985 986–1081 1082–1177 1178–1273 1274–1369
SAT Scores 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 457.5 553.5 649.5 745.5 841.5 937.5 1033.5 1129.5 1225.5 1321.5
Class
Max Min 1359 410 94.9 ⇒ 95 Number of classes 10
Relative frequency
44. (a) Class width
SAT scores
f
f 50
N1
(b) 48%, because the sum of the relative frequencies for the last four classes is 0.48. (c) 698, because the sum of the relative frequencies for the last seven classes is 0.88.
8 7 6 5 4 3 2 1
6
5
5
4
4 3 2
5
8
11
14
1.5
Data
3 2 1
1 2
Histogram (20 Classes)
Histogram (10 Classes)
Frequency
Frequency
Histogram (5 Classes)
Frequency
45.
5.5
9.5 13.5 17.5
Data
1 3 5 7 9 11 13 15 17 19
Data
In general, a greater number of classes better preserves the actual values of the data set, but is not as helpful for observing general trends and making conclusions. When choosing the number of classes, an important consideration is the size of the data set. For instance, you would not want to use 20 classes if your data set contained 20 entries. In this particular example, as the number of classes increases, the histogram shows more fluctuation. The histograms with 10 and 20 classes have classes with zero frequencies. Not much is gained by using more than five classes. Therefore, it appears that five classes would be best.
2.2
MORE GRAPHS AND DISPLAYS
2.2 Try It Yourself Solutions 1a. 1 2 3 4 5 6 7 8
b. Key: 1 7 17 1 758855 2 76898979875346250112141 3 9977886476555145982522233324110421210 4 986878648546671154532283040530 5 945435590235705 6 85133110 7 8 9
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CHAPTER 2
 DESCRIPTIVE STATISTICS
c. Key: 1 7 17 1 2 3 4 5 6 7 8
555788 01111223445566777888999 0011111222222233344445555566777888999 000112233344445555666677888889 00233445555799 01113358 9
d. It seems that most teams scored under 54 touchdowns.
2ab. Key: 1 7 17
1 1 555788 2 0111122344 2 5566777888999 3 001111122222223334444 3 5555566777888999 4 00011223334444 4 5555666677888889 5 0023344 5 5555799 6 011133 6 58 7 7 8 8 9 3a. Use number of touchdowns for the horizontal axis. b.
Touchdowns Scored
10
20
30
40
50
60
70
80
90
Number of Touchdowns
c. It appears that a large percentage of teams scored under 50 touchdowns. 4a.
b. Motor Vehicle Occupants
Vehicle type
Killed (frequency)
Relative frequency
Central angle
Cars Trucks Motorcycles Other
22,423 10,216 2,227 425
0.64 0.29 0.06 0.01
0.64360 230 0.29360 104 0.06360 22 0.01360 4
f
35,291
f
n1
360
Killed in 1995 Trucks 29%
6%
Motorcycles Other 1%
Cars 64%
c. As a percentage of total vehicle deaths, car deaths decreased by 15%, truck deaths increased by 8%, and motorcycle deaths increased by 6%. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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CHAPTER 2
5a.
Cause

DESCRIPTIVE STATISTICS
b.
Frequency, f 14,668 9,728 7,792 5,733 4,649
Auto dealers Auto repairs Home furnishing Computer sales Dry cleaning
Frequency
Auto Dealers Auto Repair Home Furnishing Computer Sales Dry Cleaning
Causes of BBB Complaints 16,000 14,000 12,000 10,000 8,000 6,000 4,000 2,000
Cause
c. It appears that the auto industry (dealers and repair shops) account for the largest portion of complaints filed at the BBB. 6ab. Salary (in dollars)
Salaries 50,000 45,000 40,000 35,000 30,000 25,000 20,000 2
4
6
8
10
Length of employment (in years)
c. It appears that the longer an employee is with the company, the larger his/her salary will be. Cellular Phone Bills 52 50 48 46 44 42 40 38
1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005
Average bill (in dollars)
7ab.
Year
c. It appears that the average monthly bill for cellular telephone subscribers decreased significantly from 1995 to 1998, then increased from 1998 to 2004.
2.2 EXERCISE SOLUTIONS 1. Quantitative: StemandLeaf Plot, Dot Plot, Histogram, Time Series Chart, Scatter Plot Qualitative: Pie Chart, Pareto Chart 2. Unlike the histogram, the stemandleaf plot still contains the original data values. However, some data are difficult to organize in a stemandleaf plot. 3. Both the stemandleaf plot and the dot plot allow you to see how data are distributed, determine specific data entries, and identify unusual data values. 4. In the pareto chart, the height of each bar represents frequency or relative frequency and the bars are positioned in order of decreasing height with the tallest bar positioned to the left. 5. b
6. d
7. a
8. c
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CHAPTER 2
 DESCRIPTIVE STATISTICS
9. 27, 32, 41, 43, 43, 44, 47, 47, 48, 50, 51, 51, 52, 53, 53, 53, 54, 54, 54, 54, 55, 56, 56, 58, 59, 68, 68, 68, 73, 78, 78, 85 Max: 85
Min: 27
10. 12.9, 13.3, 13.6, 13.7, 13.7, 14.1, 14.1, 14.1, 14.1, 14.3, 14.4, 14.4, 14.6, 14.9, 14.9, 15.0, 15.0, 15.0, 15.1, 15.2, 15.4, 15.6, 15.7, 15.8, 15.8, 15.8, 15.9, 16.1, 16.6, 16.7 Max: 16.7
Min: 12.9
11. 13, 13, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 17, 18, 19 Max: 19
Min: 13
12. 214, 214, 214, 216, 216, 217, 218, 218, 220, 221, 223, 224, 225, 225, 227, 228, 228, 228, 228, 230, 230, 231, 235, 237, 239 Max: 239
Min: 214
13. AnheuserBusch is the top sports advertiser spending approximately $190 million. Honda spends the least. (Answers will vary.) 14. The value of the stock portfolio has increased fairly steadily over the past five years with the greatest increase happening between 2003 and 2006. (Answers will vary.) 15. Tailgaters irk drivers the most, while too cautious drivers irk drivers the least. (Answers will vary.) 16. The most frequent incident occurring while driving and using a cell phone is swerving. Twice as many people “sped up” than “cut off a car.” (Answers will vary.)
17. Key: 6 7 67 6 7 8 9
78 35569 002355778 01112455
Most grades of the biology midterm were in the 80s or 90s.
18. Key: 4 0 40 4 5 6 7 8
0799 01246899 1237 13689 0447
It appears that most of the world’s richest people are over 49 years old. (Answers will vary.)
19. Key: 4 3 4.3 4 5 6 7 8
39 18889 48999 002225 01
It appears that most ice had a thickness of 5.8 centimeters to 7.2 centimeters. (Answers will vary.) © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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DESCRIPTIVE STATISTICS
20. Key: 17 5 17.5 16 17 18 19 20
48 113455679 13446669 0023356 18
It appears that most farmers charge 17 to 19 cents per pound of apples. (Answers will vary.) 21.
Advertisements
150
250
350
450
550
650
750
850
Number of ads
It appears that most of the 30 people from the U.S. see or hear between 450 and 750 advertisements per week. (Answers will vary.) 22.
It appears that the lifespan of a fly tends to be between 8 and 11 days. (Answers will vary.)
Housefly Life Spans
4 5 6 7 8 9 10 11 12 13 14
Life span (in days)
23. Category
Countries in the United Nations
Frequency
Relative frequency
Angle
23 12 43 14 53 47
0.12 0.06 0.22 0.07 0.28 0.25
0.12360 43 0.06360 23 0.22360 81 0.07360 26 0.28360 99 0.25360 88
f 192
n1
North America South America Europe Oceania Africa Asia
North America 12% Asia 25%
Oceania 7% South America 6%
Europe 22%
Africa 28%
f
Most countries in the United Nations come from Africa and the least amount come from South America. (Answers will vary.) 24. Category Science, aeronautics, and exploration Exploration capabilities Inspector General
Budget frequency
Relative frequency
Angle
10,651
0.6343
0.6343360 228
6,108 34
0.3637 0.0020
0.3637360 131 0.0020360 0.7
f
f 16,793 N 1 2007 NASA Budget Exploration capabilities 36.37%
Inspector General 0.20%
Science, aeronautics, and exploration 63.43%
It appears that 63.4% of NASA’s budget went to science, aeronautics, and exploration. (Answers will vary.) © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2
25.
Other
It appears that the biggest reason for baggage delay comes from transfer baggage mishandling. (Answers will vary.)
Spaceweight restriction Loading/offloading error Arrival station mishandling
70 60 50 40 30 20 10
Transfer baggage mishandling Failure to load at originating airport
Percents
Airline Baggage Delay
 DESCRIPTIVE STATISTICS
Reason for delay
26.
It appears that Boise, ID and Denver, CO have the same UV index. (Answers will vary.)
Denver, CO
Boise, ID
Atlanta, GA
Concord, NH
Miami, FL
UV index
Ultraviolet Index 10 8 6 4 2
City
27. Hourly wage (in dollars)
Hourly Wages 14.00
It appears that hourly wage increases as the number of hours worked increases. (Answers will vary.)
13.00 12.00 11.00 10.00 9.00 25 30 35 40 45 50
Hours
28. Avg. teacher’s salary
Teachers’ Salaries 55 50 45 40 35 30 25 13
15
17
19
It appears that a teacher’s average salary decreases as the number of students per teacher increases. (Answers will vary.)
21
Students per teacher
29.
Ultraviolet Index
UV index
10 8
Of the period from June 14 –23, 2001, in Memphis, TN, the ultraviolet index was highest from June 16–21. (Answers will vary.)
6 4 2 14 15 16 17 18 19 20 21 22 23
Date in June
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25
26

CHAPTER 2
30.
DESCRIPTIVE STATISTICS
Temperature (in °F)
Daily High Temperatures in May
It appears that it was hottest from May 7 to May 11. (Answers will vary.)
90 88 86 84 82 80 78 76 2
4
6
8
10 12
Day of the month
31.
It appears the price of eggs peaked in 2003. (Answers will vary.)
1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005
Price of Grade A eggs (in dollars per dozen)
Price of Grade A Eggs 1.60 1.50 1.40 1.30 1.20 1.10 1.00 0.90 0.80
Year
32.
It appears that the greatest increases in dollars per pound was 2002 to 2003. (Answers will vary.)
2.10 1.90 1.70 1.50 1.30 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005
Price of Beef (in dollars per pound)
Price of 100% Ground Beef 2.30
Year
33. (a) When data are taken at regular intervals over a period of time, a time series chart should be used. (Answers will vary.) (b) Sales (thousands of dollars)
Sales for Company A 130 120 110 100 90 1st
2nd 3rd
4th
Quarter
34. (a) The pie chart should be displaying all four quarters, not just the first three. (b)
Sales for Company B
4th quarter 20%
1st quarter 20%
2nd quarter 15% 3rd quarter 45%
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CHAPTER 2
 DESCRIPTIVE STATISTICS
35. (a) At law firm A, the lowest salary was $90,000 and the highest was $203,000; at law firm B, the lowest salary was $90,000 and the highest salary was $190,000. (b) There are 30 lawyers at law firm A and 32 lawyers at law firm B. (c) At Law Firm A, the salaries tend to be clustered at the far ends of the distribution range and at Law Firm B, the salaries tend to fall in the middle of the distribution range. 36. (a) At the 3:00 P.M. class, the youngest participant is 35 years old and the oldest participant is 85 years old. At the 8:00 P.M. class, the youngest participant is 18 years old and the oldest participant is 71 years old. (b) In the 3:00 P.M. class, there are 26 particpants and in the 8:00 P.M. class, there are 30 particpants. (c) The participants in each class are clustered at one of the ends of their distribution range. The 3:00 P.M. class mostly has particpants over 50 and the 8:00 P.M. class mostly has participants under 50. (Answers will vary.)
2.3
MEASURES OF CENTRAL TENDENCY
2.3 Try It Yourself Solutions 1a. x 578 b. x
x 578 41.3 n 14
c. The mean age of an employee in a department is 41.3 years. 2a. 18 18, 19, 19, 19, 20, 21, 21, 21, 21, 23, 24, 24, 26, 27, 27, 29, 30, 30, 30, 33, 33, 34, 35, 38 b. median middle entry 24 c. The median age for the sample of fans at the concert is 24. 3a. 70, 80, 100, 130, 140, 150, 160, 200, 250, 270 b. median mean of two middle entries 140, 150 145 c. The median price of the sample of MP3 players is $145. 4a. 0, 0, 1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 7, 9, 10, 12, 12, 13, 13, 13, 13, 13, 15, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 22, 23, 23, 24, 24, 25, 25, 26, 26, 26, 29, 33, 36, 37, 39, 39, 39, 39, 40, 40, 41, 41, 41, 42, 44, 44, 45, 47, 48, 49, 49, 49, 51, 53, 56, 58, 58, 59, 60, 67, 68, 68, 72 b. The age that occurs with the greatest frequency is 13 years old. c. The mode of the ages is 13 years old. 5a. “Yes” occurs with the greatest frequency (171). b. The mode of the responses to the survey is “Yes”.
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CHAPTER 2
6a. x

DESCRIPTIVE STATISTICS
x 410 21.6 n 19
median 21 mode 20 b. The mean in Example 6 x 23.8 was heavily influenced by the age 65. Neither the median nor the mode was affected as much by the age 65. 7ab. Source Test Mean Midterm Final Computer Lab Homework
Score, x
Weight, w
86 96 98 98 100
0.50 0.15 0.20 0.10 0.05
x
ⴢw
830.50 43.0 960.15 14.4 980.20 19.6 980.10 9.8 1000.05 5.0
w 1.00 x w 91.8 c. x
x w 91.8 91.8 w 1.00
d. The weighted mean for the course is 91.8. So, you did get an A. 8abc. Class
Midpoint, x
Frequency, f
15–24 25–34 35–44 45–54 55–64 65–74 75–84 85–94
19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5
16 34 30 23 13 2 0 1 N 119
d.
x
ⴢf
312 1003 1185 1138.5 773.5 139 0 89.5
x f 4640.5
x f 4640.5 39.0 N 119
The average number of touchdowns is approximately 39.0.
2.3 EXERCISE SOLUTIONS 1. True. 2. False. Not all data sets must have a mode. 3. False. All quantitative data sets have a median. 4. False. The mode is the only measure of central tendency that can be used for data at the nominal level of measurement. 5. False. When each data class has the same frequency, the distribution is uniform. 6. False. When the mean is greater than the median, the distrubution is skewed right.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
7. Answers will vary. A data set with an outlier within it would be an example. For instance, the mean of the prices of existing home sales tends to be “inflated” due to the presence of a few very expensive homes. 8. Any data set that is symmetric has the same median and mode. 9. Skewed right because the “tail” of the distribution extends to the right. 10. Symmetric because the left and right halves of the distribution are approximately mirror images. 11. Uniform because the bars are approximately the same height. 12. Skewed left because the tail of the distribution extends to the left. 13. (11), because the distribution values range from 1 to 12 and has (approximately) equal frequencies. 14. (9), because the distribution has values in the thousands of dollars and is skewed right due to the few executives that make a much higher salary than the majority of the employees. 15. (12), because the distribution has a maximum value of 90 and is skewed left due to a few students scoring much lower than the majority of the students. 16. (10), because the distribution is rather symmetric due to the nature of the weights of seventh grade boys. 17. x
x 81 6.2 n 13
5 5 5 5 5 5 6 6 7 8 9 9 middle value ⇒ median 6 mode 5 18. x
(occurs 6 times)
x 252 25.2 n 10
19 20 21 22 22 23 25 30 35 35 two middle values ⇒ median mode 22, 35 19. x
22 23 22.5 2
(occurs 2 times each)
x 32 4.57 n 7
3.7 4.0 4.8 4.8 4.8 4.8 5.1 middle value ⇒ median 4.8 mode 4.8 20. x 154
(occurs 4 times)
x 2004 200.4 n 10 171
173
181
184
188
203
235
240
275
two middle values ⇒ median
184 188 186 2
mode none The mode cannot be found because no data points are repeated.
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29
30
CHAPTER 2
21. x

DESCRIPTIVE STATISTICS
x 661.2 20.66 n 32
10.5, 13.2, 14.9, 16.2, 16.7, 16.9, 17.6, 18.2, 18.6, 18.8, 18.8, 19.1, 19.2, 19.6, 19.8, 19.9, 20.2, 20.7, 20.9, 22.1, 22.1, 22.2, 22.9, 23.2, 23.3, 24.1, 24.9, 25.8, 26.6, 26.7, 26.7, 30.8 two middle values ⇒ median mode 18.8, 22.1, 26.7 22. x
19.9 20.2 20.05 2
(occurs 2 times each)
x 1223 61.2 n 20
12 18 26 28 31 33 40 44 45 49 61 63 75 80 80 89 96 103 125 125 49 61 two middle values ⇒ median 55 2 mode 80, 125 The modes do not represent the center of the data set because they are large values compared to the rest of the data. 23. x not possible nominal data median not possible nominal data mode “Worse” The mean and median cannot be found because the data are at the nominal level of measurement. 24. x not possible (nominal data) median not possible (nominal data) mode “Watchful” The mean and median cannot be found because the data are at the nominal level of measurement. 25. x
x 1194.4 170.63 n 7
155.7, 158.1, 162.2, 169.3, 180, 181.8, 187.3 middle value ⇒ median 169.3 mode none The mode cannot be found because no data points are repeated. 26. x not possible (nominal data) median not possible (nominal data) mode “Domestic” The mean and median cannot be found because the data are at the nominal level of measurement.
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CHAPTER 2
27. x
 DESCRIPTIVE STATISTICS
x 226 22.6 n 10
14, 14, 15, 177, 18, 20, 22, 25, 40, 41 two middle values ⇒ median mode 14 28. x
18 20 19 2
(occurs 2 times)
x 83 16.6 n 5
1, 10, 15, 25.5, 31.5 middle value ⇒ median 15 mode none The mode cannot be found because no data points are repeated. 29. x
x 197.5 14.11 n 14
1.5, 2.5, 2.5, 5, 10.5, 11, 13, 15.5, 16.5, 17.5, 20, 26.5, 27, 28.5 two middle values ⇒ median mode 2.5 30. x
13 15.5 14.25 2
(occurs 2 times)
x 3455 314.1 n 11
25, 35, 93, 110, 356, 374, 380, 445, 458, 480, 699 middle value ⇒ median 374 mode none The mode cannot be found because no data points are repeated. 31. x
x 578 41.3 n 14
10, 12, 21, 24, 27, 37, 38, 41, 45, 45, 50, 57, 65, 106 two middle values ⇒ median mode 4.5 32. x
38 41 39.5 2
(occurs 2 times)
x 29.9 2.49 n 12
0.8, 1.5, 1.6, 1.8, 2.1, 2.3, 2.4, 2.5, 3.0, 3.9, 4.0, 4.0 two middle values ⇒ median mode 4.0
2.3 2.4 2.35 2
(occurs 2 times)
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31
32
CHAPTER 2
33. x

DESCRIPTIVE STATISTICS
x 292 19.5 n 15
5, 8, 10, 15, 15, 15, 17, 20, 21, 22, 22, 25, 28, 32, 37 middle value ⇒ median 20 mode 15 34. x
(occurs 3 times)
x 2987 213.4 n 14
205, 208, 210, 212, 212, 214, 214, 214, 215, 215, 217, 217, 217, 217 two middle values ⇒ median mode 217
214 214 214 2
(occurs 4 times)
35. A mode (data entry that occurred most often) B median (left of mean in skewedright dist.) C mean (right of median in skewedright dist.) 36. A mean (left of median in skewedleft dist.) B median (right of mean in skewedleft dist.) C mode (data entry that occurred most often) 37. Mode because the data is nominal.
38. Mean because the data are symmetric.
39. Mean because the data does not contain outliers. 40. Median because the data are skewed. 41.
Source Homework Quiz Project Speech Final Exam
Score, x
Weight, w
85 80 100 90 93
0.05 0.35 0.20 0.15 0.25
xⴢw
x
43.
Score, x
Weight, w
xⴢw
MBAs Bas
$45,500 $32,000
8 17
45,5008 364,000 32,00017 544,000
Balance, x $523 $2415 $250
x
xⴢw
24 2 4
52324 12,552 24152 4830 2504 1000 x
x w 908,000 $36,320 w 25
x
x w 18,382 $612.73 w 30
w 908,000
Days, w
w 30
x
w 89
Source
w 25
x w 89 89 w 1
850.05 4.25 800.35 28 1000.20 20 900.15 13.5 930.25 23.25
w 1
42.
x
w 18,382
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CHAPTER 2
44.
Balance, x
Days, w
$759 $1985 $1410 $348
46.
Credits, w
B B A D C
3 3 4 1 2
3 3 4 2 3
33 9 33 9 44 16 12 2 23 6
w 15
x w 42
Score, x
Weight, w
xⴢw
85 81 90
9 13 5
859 765 8113 1053 905 450
x
47.
Midpoint, x
Frequency, f
61 64 67 70
4 5 8 1 n 18
48.
Midpoint, x
Frequency, f
64 67 70 73 76
3 6 7 4 3 x
49.
x
ⴢw
Points, x
Engineering Business Math
x w 30,448 $982.19 w 31
x
x w 42 2.8 w 15
x
x w 2268 84 w 27
x
x f 1170 65 inches n 18
w 30,448
x
Grade
Source
x
75915 11,385 19855 9925 14105 7050 3486 2080
15 5 5 6 w 31
45.
ⴢw
x
f 23
w 27
w 2268
ⴢf
x
614 244 645 320 678 536 701 70 x
f 1170 x
ⴢf
x
x f 1604 69.7 inches n 23
x
x f 14,196.5 35.8 years old n 397
643 192 676 402 707 490 734 292 763 228 x
f 1604
Midpoint, x
Frequency, f
xⴢf
4.5 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5
55 70 35 56 74 42 38 17 10
4.555 247.5 14.570 1015 24.535 857.5 34.556 1932 44.574 3293 54.542 2289 64.538 2451 74.517 1266.5 84.510 845
n 397
 DESCRIPTIVE STATISTICS
x
f 14,196.5
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33
34
50.
CHAPTER 2

DESCRIPTIVE STATISTICS
Midpoint, x
Frequency, f
3 8 13 18 23 28 33 38 43
12 26 20 7 11 7 4 4 1
x
x
x f 1406 15.3 minutes n 92
312 36 826 208 1320 260 187 126 2311 253 287 196 334 132 384 152 431 43 x
n 92
51. Class width
ⴢf
f 1406
Max Min 14 3 1.83 ⇒ 2 Number of classes 6 Hospitalization
Frequency, f
13.5
9.5
11.5
f 20
7.5
3 8 4 2 2 1
5.5
3.5 5.5 7.5 9.5 11.5 13.5
8 7 6 5 4 3 2 1
3.5
3–4 5–6 7–8 9–10 11–12 13–14
Midpoint, x
Frequency
Class
Days hospitalized
Shape: Positively skewed
Class
Max Min 297 127 34 Number of classes 5
Midpoint, x
Frequency, f
144 179 214 249 284
9 8 3 3 1
127–161 162–196 197–231 232–266 267–301
Hospital Beds
Frequency
52. Class width
9 8 7 6 5 4 3 2 1 144 179 214 249 284
Number of beds
f 24
Shape: Positively skewed Max Min 76 62 2.8 ⇒ 3 Number of classes 5
Class
Midpoint, x
Frequency, f
62–64 65–67 68–70 71–73 74–76
63 66 69 72 75
3 7 9 8 3 f 30
Heights of Males Frequency
53. Class width
9 8 7 6 5 4 3 2 1 63
66
69
72
75
Heights (to the nearest inch)
Shape: Symmetric
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CHAPTER 2
54. Class width
61 Max Min 0.8333 ⇒ 1 Number of classes 6
 DESCRIPTIVE STATISTICS
Results of Rolling SixSided Die
Frequency, f
1 2 3 4 5 6
6 5 4 6 4 5
Frequency
6
Class
5 4 3 2 1 1
2
3
4
5
6
Number rolled
f 30
Shape: Uniform 55. (a) x
x 36.03 6.005 n 6
5.59, 5.99, 6, 6.02, 6.03, 6.4 two middle values ⇒ median (b) x
6 6.02 6.01 2
x 35.67 5.945 n 6
5.59, 5.99, 6, 6.02, 6.03, 6.4 two middle values ⇒ median
6 6.02 6.01 2
(c) mean 56. (a) x
x 815.4 42.92 n 19
7.8, 8.2, 12.6, 12.6, 14.4, 17.8, 19.2, 21.3, 23, 24.2, 24.7, 31.1, 32.5, 41.3, 45.4, 55.2, 59.6, 134.2, 230.3 middle value ⇒ median 24.2 (b) x
x 585.1 32.51 n 18
7.8, 8.2, 12.6, 12.6, 14.4, 17.8, 19.2, 21.3, 23, 24.2, 24.7, 31.1, 32.5, 41.3, 45.4, 55.5, 59.6, 134.2 two middle values ⇒ median
23 24.2 23.6 2
(c) mean 57. (a) x
x 3222 358 n 9
147, 177, 336, 360, 375, 393, 408, 504, 522 middle value ⇒ median 375
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35
36
CHAPTER 2
(b) x

DESCRIPTIVE STATISTICS
x 9666 1074 n 9
441, 531, 1008, 1080, 1125, 1179, 1224, 1512, 1566 middle value ⇒ median 1125 (c) x from part (b) is 3 times x from part (a). Median from part (b) is 3 times median from part (a). (d) Multiply part (b) answers by 12. 58. (a) Mean should be used because Car A has the highest mean of the three. (b) Median should be used because Car B has the highest median of the three. (c) Mode should be used because Car C has the highest mode of the three. 59. Car A because it has the highest midrange of the three. Car A: Midrange
34 28 32 2
Car B: Midrange
31 29 30 2
Car C: Midrange
32 28 30 2
60. (a) x 49.2, median 46.5
(b) Key: 3 6 36 1 2 3 4 5 6 7 8 9
(c) Positively skewed
13 28 6667778 13467 1113 1234 2246 5 0
median mean
61. (a) Order the data values. 11 47
13 51
22 51
28 51
36 53
36 61
36 62
37 63
37 64
37 72
38 72
41 74
43 76
44 85
46 90
Delete the lowest 10%, smallest 3 observations 11, 13, 22. Delete the highest 10%, largest 3 observations 76, 85, 90. Find the 10% trimmed mean using the remaining 24 observations. 10% trimmed mean 49.2
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CHAPTER 2
 DESCRIPTIVE STATISTICS
(b) x 49.2 median 46.5 mode 36, 37, 51 midrange 50.5 (c) Using a trimmed mean eliminates potential outliers that may affect the mean of all the observations.
2.4
MEASURES OF VARIATION
2.4 Try It Yourself Solutions 1a. Min 23 or $23,000 and Max 58 or $58,000 b. Range max min 58 23 35 or $35,000 c. The range of the starting salaries for Corporation B is 35 or $35,000 (much larger than range of Corporation A). 2a. b.
x 415 41.5 or $41,500 10
Salary, x (1000s of dollars) 23 29 32 40 41 41 49 50 52 58
Deviation, x ⴚ (1000s of dollars) 23 29 32 40 41 41 49 50 52 58
x 415
41.5 41.5 41.5 41.5 41.5 41.5 41.5 41.5 41.5 41.5
18.5 12.5 9.5 1.5 0.5 0.5 7.5 8.5 10.5 16.5
x 0
3ab. 41.5 or $41,500 Salary, x
xⴚ
x ⴚ 2
23 29 32 40 41 41 49 50 52 58
18.5 12.5 9.5 1.5 0.5 0.5 7.5 8.5 10.5 16.5
18.52 342.25 12.52 156.25 9.52 90.25 1.52 2.25 0.52 0.25 0.52 0.25 7.52 56.25 8.52 72.25 10.52 110.25 16.52 272.25
x 415
x 0
x
2
1102.5
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37
38
CHAPTER 2
c. 2

DESCRIPTIVE STATISTICS
x 2 1102.5 110.25 N 10
d. 2
10.5 or $10,500 1102.5 10
e. The population variance is 110.3 and the population standard deviation is 10.5 or $10,500. 4a.
Salary, x
xⴚx
x ⴚ x 2
23 29 32 40 41 41 49 50 52 58
18.5 12.5 9.5 1.5 0.5 0.5 7.5 8.5 10.5 16.5
342.25 156.25 90.25 2.25 0.25 0.25 56.25 72.25 110.25 272.25
x 415
x x 0
x x
2
1102.5
SSx x x 2 1102.5 b. s2
x x2 1102.5 122.5 n 1 9
c. s s2 122.5 11.1 or $11,100 5a. (Enter data in computer or calculator) b. x 37.89,
s ⴝ 3.98
6a. 7, 7, 7, 7, 7, 13, 13, 13, 13, 13 b.
x
xⴚ
x ⴚ 2
7 7 7 7 7 13 13 13 13 13
7 10 3 7 10 3 7 10 3 7 10 3 7 10 3 13 10 3 13 10 3 13 10 3 13 10 3 13 10 3
32 9 32 9 32 9 32 9 32 9 32 9 32 9 32 9 32 9 32 9
x 100
x 0 x
x 100 10 N 10
x N 9010 2
2
90
9 3
7a. 64 61.25 2.75 1 standard deviation b. 34% c. The estimated percent of the heights that are between 61.25 and 64 inches is 34%.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
8a. 31.6 219.5 7.4 > 0 b. 31.6 219.5 70.6 c. 1
1 1 1 1 2 1 0.75 k2 2 4
At least 75% of the data lie within 2 standard deviations of the mean. At least 75% of the population of Alaska is between 0 and 70.6 years old. 9a.
x
f
0 1 2 3 4 5 6
10 19 7 7 5 1 1
xf
010 0 119 19 27 14 37 21 45 20 51 5 61 6
xf 85
n 50
b. x c.
xf 85 1.7 n 50 x ⴚ x 2
x ⴚ x 2 ⴢ f
1.70 2.8900 0.702 0.4900 0.302 0.0900 1.302 1.6900 2.302 5.2900 3.302 10.9800 4.302 18.4900
2.890010 28.90 0.490019 9.31 0.09007 0.63 1.69007 11.83 5.29005 26.45 10.98001 10.89 18.49001 18.49
xⴚx 0 1.7 1.70 1 1.7 0.70 2 1.7 0.30 3 1.7 1.30 4 1.7 2.30 5 1.7 3.30 6 1.7 4.30
2
x x f 106.5 2
d. s 10a.
Class 0–99 100–199 200–299 300–399 400–499 500+
b. x
x x2f n 1
106.5 49
2.17 1.5
x
f
xf
49.5 149.5 249.5 349.5 449.5 650.0
380 230 210 50 60 70
49.5380 18,810 149.5230 34,385 249.5210 52,395 349.550 17,475 449.560 26,970 650.070 45,500
n 1000
xf 195,535
xf 195,535 195.5 n 1000
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39
40
CHAPTER 2
c.

DESCRIPTIVE STATISTICS
xⴚx
x ⴚ x 2
x ⴚ x 2 ⴢ f
49.5 195.5 146 149.5 195.5 46 249.5 195.5 54 349.5 195.5 154 449.5 195.5 254 650 195.5 454.5
1462 21,316 462 2116 542 2916 1542 23,716 2542 64,516 454.52 206,570.25
21,316380 8,100,080 2116230 486,680 2916210 612,360 23,71650 1,185,800 64,51660 3,870,960 206,570.2570 14,459,917.5
x x f 28,715,797.5 2
d. s
x x2f n1
28,715,797.5 999
28,744.542 169.5
2.4 EXERCISE SOLUTIONS 1. Range Max Min 12 4 8
x 79 7.9 N 10 x ⴚ 2
x
xⴚ
12 9 7 5 7 8 10 4 11 6
12 7.9 4.1 9 7.9 1.1 7 7.9 0.9 5 7.9 2.9 7 7.9 0.9 8 7.9 0.1 10 7.9 2.1 4 7.9 3.9 11 7.9 3.1 6 7.9 1.9
4.12 16.81 1.12 1.21 0.92 0.81 2.92 8.41 0.92 0.81 0.12 0.01 2.12 4.41 3.92 15.21 3.12 9.61 1.92 3.61
x 79
x 0
x 2 60.9
2
x 2 60.9 6.09 6.1 N 10
x N
2
6.09 2.5
2. Range Max Min 24 14 10
x 264 18.9 N 14
x 15 24 17 19 20 18 20 16 21 23 17 18 22 14 x 264
x ⴚ 2
xⴚ 15 18.9 24 18.9 17 18.9 19 18.9 20 18.9 18 18.9 20 18.9 16 18.9 21 18.9 23 18.9 17 18.9 18 18.9 22 18.9 14 18.9
3.9 5.1 1.9 0.1 1.1 0.9 1.1 2.9 2.1 4.1 1.9 0.9 3.1 4.9
x 0
3.92 15.21 5.12 26.01 1.92 3.61 0.12 0.01 1.12 1.21 0.92 0.81 1.12 1.21 2.92 8.41 2.12 4.41 4.12 16.81 1.92 3.61 0.92 0.81 3.12 9.61 4.92 24.01 x 2 115.74
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CHAPTER 2
x 2 115.74 8.27 8.3 N 14
2
 DESCRIPTIVE STATISTICS
x 2 8.27 2.9 N
3. Range Max Min 18 6 12 x 107 x 11.9 n 9 x
x ⴚ x
17 8 13 18 15 9 10 11 6
17 11.9 5.1 8 11.9 3.9 13 11.9 1.1 18 11.9 6.1 15 11.9 3.1 9 11.9 2.9 10 11.9 1.9 11 11.9 0.9 6 11.9 5.9
x 107
x x 0
s2 s
x ⴚ x 2 5.12 26.01 3.92 15.21 1.12 1.21 6.12 37.21 3.12 9.61 2.92 8.41 1.92 3.61 0.92 0.81 5.92 34.81 x x 2 136.89
x x2 136.89 17.1 n1 91
nx 1x
2
17.1 4.1
4. Range Max Min 28 7 21 x
x 238 18.3 n 13 x
x ⴚ x
x ⴚ x 2
28 25 21 15 7 14 9 27 21 24 14 17 16
28 18.3 9.7 25 18.3 6.7 21 18.3 2.7 15 18.3 3.3 7 18.3 11.3 14 18.3 4.3 9 18.3 9.3 27 18.3 8.7 21 18.3 2.7 24 18.3 5.7 14 18.3 4.3 17 18.3 1.3 16 18.3 2.3
9.72 94.09 6.72 44.89 2.72 7.29 3.32 10.89 11.32 127.69 4.32 18.49 9.32 86.49 8.72 75.69 2.72 7.29 5.72 32.49 4.32 18.49 1.32 1.69 2.32 5.29
x 238
x x 0
x x2 530.77
s2 s
x x2 530.77 44.23 44.2 n1 13 1
nx 1x
2
44.23 6.7
5. Range Max Min 96 23 73 6. Range Max Min 34 24 10
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41
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CHAPTER 2

DESCRIPTIVE STATISTICS
7. The range is the difference between the maximum and minimum values of a data set. The advantage of the range is that it is easy to calculate. The disadvantage is that it uses only two entries from the data set. 8. The deviation, x , is the difference between an observation, x, and the mean of the data, .The sum of the deviations is always zero. 9. The units of variance are squared. Its units are meaningless. (Ex: dollars2) 10. The standard deviation is the positive square root of the variance. Because squared deviations can never be negative, the standard deviation and variance can never be negative.
7, 7, 7, 7, 7 → n 5 x7 s0 11. (a) Range Max Min 45.6 21.3 24.3 (b) Range Max Min 65.6 21.3 44.3 (c) The range has increased substantially. 12. 3, 3, 3, 7, 7, 7 → n 6
5 s2 13. Graph (a) has a standard deviation of 24 and graph (b) has a standard deviation of 16 because graph (a) has more variability. 14. Graph (a) has a standard deviation of 2.4 and graph (b) has a standard deviation of 5. Graph (b) has more variability. 15. When calculating the population standard deviation, you divide the sum of the squared deviations by N, then take the square root of that value. When calculating the sample standard deviation, you divide the sum of the squared deviations by n 1, then take the square root of that value. 16. When given a data set, one would have to determine if it represented the population or was a sample taken from a population. If the data are a population, then is calculated. If the data are a sample, then s is calculated. 17. Company B. Due to the larger standard deviation in salaries for company B, it would be more likely to be offered a salary of $33,000. 18. Player B. Due to the smaller standard deviation in number of strokes, player B would be the more consistent player.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
19. (a) Los Angeles: range Max Min 35.9 18.3 17.6 x
x ⴚ x
x ⴚ x 2
20.2 26.1 20.9 32.1 35.9 23.0 28.2 31.6 18.3
6.06 0.16 5.36 5.84 9.64 3.64 1.94 5.34 7.96
36.67 0.02 28.68 34.16 93.02 10.60 3.78 28.56 63.29 x x 2 298.78
x 236.3
x
x 236.3 26.26 n 9
s2
x x2 298.78 37.35 (n 1) 8
s s2 6.11 Long Beach: range Max Min 26.9 18.2 8.7 x
x ⴚ x
x ⴚ x 2
20.9 18.2 20.8 21.1 26.5 26.9 24.2 25.1 22.2
1.98 4.68 2.08 1.78 3.62 4.02 1.32 2.22 0.68
3.91 21.88 4.32 3.16 13.12 16.18 1.75 4.94 0.46
x 205.9
x x 2 69.72
x
x 205.9 22.88 n 9
s2
x x2 69.72 8.71 n 1 8
s s2 2.95 (b) It appears from the data that the annual salaries in Los Angeles are more variable than the salaries in Long Beach.
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20. (a) Dallas: range Max Min 34.9 16.8 18.1 x
xⴚx
x ⴚ x 2
34.9 25.7 17.3 16.8 26.8 24.7 29.4 32.7 25.5
8.92 0.28 8.68 9.18 0.82 1.28 3.42 6.72 0.48
79.61 0.08 75.30 84.23 0.68 1.63 11.71 45.19 0.23 x x 2 298.66
x 233.8
x
x 233.8 25.98 n 9
s2
x x2 298.66 37.33 n 1 8
s s2 6.11 Houston: range Max Min 31.3 18.3 13 x
xⴚx
x ⴚ x 2
25.6 23.2 26.7 27.7 25.4 26.4 18.3 26.1 31.3
0.03 2.43 1.07 2.07 0.23 0.77 7.33 0.47 5.67
0.00 5.92 1.14 4.27 0.05 0.59 53.78 0.22 32.11
x 230.7
x x 2 98.08
x
x 230.7 25.63 n 9
s2
x x2 98.08 12.26 n 1 8
s s2 3.50 (b) It appears from the data that the annual salaries in Dallas are more variable than the salaries in Houston.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
21. (a) Male: range Max Min 1328 923 405 x
x ⴚ x
1059 1328 1175 1123 923 1017 1214 1042
51.13 217.8 64.88 12.88 187.13 93.13 103.88 68.13
x ⴚ x 2 2,613.77 47,469.52 4,208.77 165.77 35,015.77 8,672.77 10,790.02 4,641.02 x x 2 113,576.88
x 8881
x
x 8881 1110.13 n 8
s2
x x2 113,576.9 16,225.3 n 1 7
s s2 127.4 Female: range Max Min 1393 841 552 x
x ⴚ x
x ⴚ x 2
1226 965 841 1053 1056 1393 1312 1222
92.50 168.50 292.50 80.50 77.50 259.50 178.50 88.50
8,556.25 28,392.25 85,556.25 6,480.25 6,006.25 67,340.25 31,862.25 7,832.25
x 9068
x x 2 242,026.00
x
x 9068 1133.50 n 8
s2
x x2 242,026 34,575.1 n 1 7
s s2 185.9 (b) It appears from the data, the SAT scores for females are more variable than the SAT scores for males.
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22. (a) Public: range Max Min 39.9 34.8 5.1 x
x ⴚ x
x ⴚ x 2
38.6 38.1 38.7 36.8 34.8 35.9 39.9 36.2
1.23 0.73 1.33 0.58 2.58 1.48 2.53 1.18
1.50 0.53 1.76 0.33 6.63 2.18 6.38 1.38 x x 2 20.68
x 299
x
x 299 37.38 n 8
s2
x x2 20.68 2.95 n 1 7
s s2 1.72 Private: range Max Min 21.8 17.6 4.2 x
x ⴚ x
x ⴚ x 2
21.8 18.4 20.3 17.6 19.7 18.3 19.4 20.8
2.26 1.14 0.76 1.94 0.16 1.24 0.14 1.26
5.12 1.29 0.58 3.75 0.03 1.53 0.02 1.59
x 156.3
x x 2 13.92
x
x 156.3 19.54 n 8
s2
x x2 13.92 1.99 n 1 7
s s2 1.41 (b) It appears from the data that the annual salaries for public teachers are more variable than the salaries for private teachers. 23. (a) Greatest sample standard deviation: (ii) Data set (ii) has more entries that are farther away from the mean. Least sample standard deviation: (iii) Data set (iii) has more entries that are close to the mean. (b) The three data sets have the same mean but have different standard deviations.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
24. (a) Greatest sample standard deviation: (i) Data set (i) has more entries that are farther away from the mean. Least sample standard deviation: (iii) Data set (iii) has more entries that are close to the mean. (b) The three data sets have the same mean, median, and mode, but have different standard deviations. 25. (a) Greatest sample standard deviation: (ii) Data set (ii) has more entries that are farther away from the mean. Least sample standard deviation: (iii) Data set (iii) has more entries that are close to the mean. (b) The three data sets have the same mean, median, and mode, but have different standard deviations. 26. (a) Greatest sample standard deviation: (iii) Data set (iii) has more entries that are farther away from the mean. Least sample standard deviation: (i) Data set (i) has more entries that are close to the mean. (b) The three data sets have the same mean and median but have different standard deviations. 27. Similarity: Both estimate proportions of the data contained within k standard deviations of the mean. Difference: The Empirical Rule assumes the distribution is bellshaped, Chebychev’s Theorem makes no such assumption. 28. You must know the distribution is bellshaped. 29. (1300, 1700) → 1500 1200, 1500 1200 → x s, x s 68% of the farms value between $1300 and $1700 per acre. 30. 95% of the data falls between x 2s and x 2s. x 2s 2400 2450 1500 x 2s 2400 2450 3300 95% of the farm values lie between $1500 and $3300 per acre. 31. (a) n 75 68%75 0.6875 51 farm values will be between $1300 and $1700 per acre. (b) n 25 68%25 0.6825 17 of these farm values will be between $1300 and $1700 per acre.
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32. (a) n 40 95% of the data lie within 2 standard deviations of the mean.
95%40 0.9540 38 farm values lie between $1500 and $3300 per acre. (b) n 60
95%20 0.9520 19 of these farm values lie between $1500 and $3300 per acre. 33. x 1500 s 200 34. x 2400 s 450
1000, 2000 are outliers. They are more than 2 standard deviations from the mean 1100, 1900. 3325, 1490 are outliers. They are more than 2 standard deviations from the mean 1500, 3300.
35. x 2s, x 2s → 1.14, 5.5 are 2 standard deviations from the mean. 1 1 1 1 2 1 0.75 ⇒ At least 75% of the eruption times lie between k2 2 4 1.14 and 5.5 minutes. 1
If n 32, at least 0.7532 24 eruptions will lie between 1.14 and 5.5 minutes. 1 1 1 1 2 1 .75 → At least 75% of the 400meter dash times lie within k2 2 4 2 standard deviations of mean.
36. 1
x 2s, x 2s → 54.97, 59.17 → At least 75% of the 400meter dash times lie between 54.97 and 59.17 seconds. 37.
38.
x
f
xf
xⴚx
x x2
x ⴚ x 2 f
0 1 2 3 4
5 11 7 10 7
0 11 14 30 28
2.08 1.08 0.08 0.93 1.93
4.31 1.16 0.01 0.86 3.71
21.53 12.71 0.04 8.56 25.94
n 40
xf 83
x x f 68.78 2
x
x 83 2.1 n 40
s
nx x1 f 68.78 39 2
1.76 1.3
x
f
xf
xⴚx
x x2
x ⴚ x 2 f
0 1 2 3
3 15 24 8
0 15 48 24
1.74 0.74 0.26 1.26
3.03 0.55 0.07 1.59
9.08 8.21 1.62 12.70
n 50
xf 87
x
xf 87 1.7 n 50
s
nx x1 f 31.62 49 2
x x f 31.62 2
0.645 0.8
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CHAPTER 2
Max Min 14 2 12 2.4 ⇒ 3 5 5 5
39. Class width Class
Midpoint, x
f
2–4 5–7 8–10 11–13 14–16
3 6 9 12 15
4 8 15 4 1
xf 12 48 135 48 15
N 32
 DESCRIPTIVE STATISTICS
xf 258
xf 258 8.1 N 32
x
x ⴚ 2
x ⴚ 2 f
5.1 2.1 0.9 3.9 6.9
26.01 4.41 0.81 15.21 47.61
104.04 35.28 12.15 50.84 47.61
x f 249.92 2
x 2 N
40. Class width Class 145–164 165–184 185–204 205–224 225–244
2.8 249.92 32
Max Min 244 145 19.8 ⇒ 20 5 5
Midpoint, x
f
xf
154.5 174.5 194.5 214.5 234.5
8 7 3 1 1
1236.0 1221.5 583.5 214.5 234.5
N 20
x f 3490.0
xf 3490 174.5 N 20
xⴚ 20 0 20 40 60
x ⴚ 2
x ⴚ 2 f
400 0 400 1600 3600
3200 0 1200 1600 3600 x 2 f 9600
x N f 9600 20 2
480 21.9
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41.
Midpoint, x
f
xf
70.5 92.5 114.5 136.5 158.5
1 12 25 10 2
70.5 1110.0 2862.5 1365.0 317.0
n 50
xf 5725
x

DESCRIPTIVE STATISTICS
xf 5725 114.5 n 50
xⴚx
x ⴚ x 2
x ⴚ x 2 f
44 22 0 22 44
1936 484 0 484 1936
1936 5808 0 4840 3872 x x 2 f 16,456
s 42.
nx 1x 16,456 49 2
Class
f
xf
0 1 2 3 4
1 9 13 5 2
0 9 26 15 8
n 30
x f 58
x
335.83 18.33
xf 58 1.9 n 30
xx
x ⴚ x 2
x ⴚ x 2 f
1.93 0.93 0.07 1.07 2.07
3.72 0.86 0.00 1.14 4.28
3.72 7.74 0.00 5.70 8.56 x x2 f 25.72
s
x x2 f n1
25.72 29
0.89 0.9
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CHAPTER 2
43.
Class
Midpoint, x
f
xf
0–4 5–13 14–17 18–24 25–34 35–44 45–64 65+
2.0 9.0 15.5 21.0 29.5 39.5 54.5 70.0
20.3 35.5 16.5 30.4 39.4 39.0 80.8 40.4
40.60 319.50 255.75 638.40 1162.30 1540.50 4403.60 2828.00
n 302.3
xf 11,188.65
x
 DESCRIPTIVE STATISTICS
xf 11,188.65 37.01 n 302.3
xx
x ⴚ x 2
x ⴚ x 2f
35.01 28.01 21.51 16.01 7.51 2.49 17.49 32.99
1225.70 784.56 462.68 256.32 56.40 6.20 305.70 1088.34
24,881.77 27,851.88 7634.22 7792.13 2222.16 241.80 24,716.72 43,968.94
x x f 139,309.56 2
s 44.
nx 1x 139,309.56 301.3 2
Midpoint, x
f
xf
5 15 25 35 45 55 65 75 85 95
11.3 12.1 12.8 16.5 18.3 15.2 17.8 13.4 7.3 1.5
56.5 181.5 320.0 577.5 823.5 836.0 1157.0 1005.0 620.5 142.5
n 126.2
x f 5720
x
462.36 21.50
xf 5720 45.32 n 126.2
x ⴚ x
x ⴚ x 2
x ⴚ x 2f
 40.32  30.32  20.32  10.32
1625.70 919.30 412.92 106.50 0.10 93.70 387.30 880.90 1574.50 2468.10
18,370.41 11,123.53 5285.12 1757.25 1.83 1424.34 6893.94 11,804.06 11,493.85 3702.15
0.32 9.68 19.68 29.68 39.68 49.68
x x f 71,856.38 2
s
xn x1 f 71,856.38 125.2 2
573.93 23.96
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45. CVheights ⫽
⭈ 100% ⫽ 72.75 ⭈ 100 ⬇ 4.7
CVweights ⫽
⭈ 100% ⫽ 187.83 ⭈ 100 ⬇ 9.8
3.44
18.47
It appears that weight is more variable than height. 46. (a)
x
x2
1059 1328 1175 1123 923 1017 1214 1042
1,121,481 1,763,584 1,380,625 1,261,129 851,929 1,034,289 1,473,796 1,085,764
兺x ⫽ 8881 兺x
⫽ 9,972,597
2
Male: s ⫽ ⫽
冪兺x
2
⫺ 关共兺x兲2兾n兴 ⫽ n⫺1
⫽ 冪113,576.875 7
x
x2
1226 965 841 1053 1056 1393 1312 1222
1,503,076 931,225 707,281 1,108,809 1,115,136 1,940,449 1,721,344 1,493,284
兺 x ⫽ 9068 兺 x
2
Female: s ⫽ ⫽
冪9,972,597 ⫺7 关共8881兲 兾8兴 2
冪16,225.268 ⬇ 127.4
⫽ 10,520,604
冪兺x
2
⫺ 关共兺x兲2兾n兴 ⫽ n⫺1
⫽ 冪242,026 7
冪10,520,604 ⫺7 关共9068兲 兾8兴 2
冪34,575.143 ⬇ 185.9
(b) The answers are the same as from Exercise 21. 47. (a) x ⬇ 41.5
s ⬇ 5.3
(b) x ⬇ 43.6
s ⬇ 5.6
(c) x ⬇ 3.5
s ⬇ 0.4
(d) By multiplying each entry by a constant k, the new sample mean is k ⭈ x and the new sample standard deviation is k ⭈ s.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
48. (a) x 41.7, s 6.0 (b) x 42.7, s 6.0 (c) x 39.7, s 6.0 (d) By adding or subtracting a constant k to each entry, the new sample mean will be x k with the sample standard deviation being unaffected. 49. (a) Male SAT Scores: x 1110.125
Female SAT Score: x 1133.5
x
x ⴚ x
x
x ⴚ x
1059 1328 1175 1123 923 1017 1214 1042
51.125 217.88 64.875 12.875 187.13 93.125 103.88 68.125
1226 965 841 1053 1056 1393 1312 1222
92.5 168.5 292.5 80.5 77.5 259.5 178.5 88.5
x x 799 ⇒
xx 799 99.9 n 8
s 127.4
Private Teachers: x 19.538
x
x ⴚ x
x
x ⴚ x
38.6 38.1 38.7 36.8 34.8 35.9 39.9 36.2
1.225 0.725 1.325 0.575 2.575 1.475 2.525 1.175
21.8 18.4 20.3 17.6 19.7 18.3 19.4 20.8
2.262 1.138 0.762 1.938 0.162 1.238 0.138 1.262
x x 11.6 ⇒
s 1.72
xx 11.6 1.45 n 8
x x 8.9
x x 11.6
xx 1238 154.8 n 8
x x 799 ⇒
s 185.9
(b) Public Teachers: x 37.375
50. 1
x x 1238
x x 799
x x 8.9 ⇒
xx 8.9 1.11 n 8
s 1.41
1 1 1 0.99 ⇒ 1 0.99 2 ⇒ k2 ⇒k k2 k 0.01
1 10 0.01
At least 99% of the data in any data set lie within 10 standard deviations of the mean. 51. (a) P
3x median 317 19 2.61; skewed left s 2.3
(b) P
3x median 332 25 4.12; skewed right s 5.1
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2.5
MEASURES OF POSITION
2.5 Try It Yourself Solutions 1a. 15, 15, 15, 17, 18, 18, 20, 21, 21, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 26, 27, 27, 27, 28, 28, 28, 29, 29, 29, 30, 30, 31, 31, 31, 31, 31, 32, 32, 32, 32, 32, 32, 32, 33, 33, 33, 34, 34, 34, 34, 35, 35, 35, 35, 35, 36, 36, 37, 37, 37, 38, 38, 38, 39, 39, 39, 40, 40, 40, 41, 41, 42, 42, 43, 43, 43, 44, 44, 44, 44, 45, 45, 45, 45, 46, 46, 46, 46, 47, 47, 48, 48, 48, 48, 49, 50, 50, 52, 53, 53, 54, 54, 55, 55, 55, 55, 57, 59, 59, 60, 61, 61, 61, 63, 63, 65, 68, 89 b. Q2 37 c. Q 1 30
Q3 47
2a. (Enter the data) b. Q 1 17
Q2 23
Q3 28.5
c. One quarter of the tuition costs is $17,000 or less, one half is $23,000 or less, and three quarters is $28,500 or less. 3a. Q 1 30
Q3 47
b. IQR Q3 Q1 47 30 17 c. The touchdowns in the middle half of the data set vary by 17 years. 4a. Min 15
Q1 30
Q 3 47 bc.
Q2 37
Max 89
Touchdowns Scored
15 10
30 37 47
20
30
40
50
89 60
70
80
90
Number of Touchdowns
d. It appears that half of the teams scored between 30 and 47 touchdowns. 5a. 50th percentile b. 50% of the teams scored 40 or fewer touchdowns. 6a. 70, 8 b. x 60: z
x 60 70 1.25 8
x 71: z
x 71 70 0.125 8
x 92: z
x 92 70 2.75 8
c. From the zscore, the utility bill of $60 is 1.25 standard deviations below the mean, the bill of $71 is 0.125 standard deviation above the mean, and the bill of $92 is 2.75 standard deviations above the mean.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
7a. Best supporting actor: 50.1, 13.9 Best supporting actress: 39.7, 1.4 b. Alan Arkin: x 72 ⇒ z
x 72 50.1 1.58 13.9
Jennifer Hudson: x 25 ⇒ z
x 25 39.7 1.05 14
c. Alan Arkin’s age is 1.58 standard deviations above the mean of the best support actors. Jennifer Hudson’s age is 1.05 standard deviations below the mean of the best supporting actresses. Neither actor’s age is unusual.
2.5 EXERCISE SOLUTIONS 1. (a)
lower half
upper half
1 2 2 4 4 5 5 5 6 6 6 7 7 7 7 8 8 8 9 9 Q1 Q1 4.5
Q2
Q3
Q2 6
Q3 7.5
(b) 1
4.5 6
7.5 9
0 1 2 3 4 5 6 7 8 9
2. (a)
lower half
upper half
1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 9 Q1 Q1 3
Q2 5
Q2
Q3
Q3 8
(b) 1
3
5
8 9
0 1 2 3 4 5 6 7 8 9
3. The soccer team scored fewer points per game than 75% of the teams in the league. 4. The salesperson sold more hardware equipment than 80% of the other sales people. 5. The student scored higher than 78% of the students who took the actuarial exam. 6. The child’s IQ is higher than 93% of the other children in the same age group. 7. True 8. False. The five numbers you need to graph a boxandwhisker plot are the minimum, the maximum, Q1, Q3, and the median Q2. 9. False. The 50th percentile is equivalent to Q2.
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10. False. Any score equal to the mean will have a corresponding zscore of zero. 11. (a) Min 10
(b) Max 20
(c) Q1 13
(d) Q2 15
(e) Q3 17
(f) IQR Q3 Q1 17 13 4
12. (a) Min 100
(b) Max 320
(c) Q1 130
(d) Q2 205
(e) Q3 270
(f) IQR Q3 Q1 270 130 140
13. (a) Min 900
(b) Max 2100
(c) Q1 1250
(d) Q2 1500
(e) Q3 1950
(f) IQR Q3 Q1 1950 1250 700
14. (a) Min 25
(b) Max 85
(c) Q1 50
(d) Q2 65
(e) Q3 70
(f) IQR Q3 Q1 70 50 20
15. (a) Min 1.9
(b) Max 2.1
(c) Q1 0.5
(d) Q2 0.1
(e) Q3 0.7
(f) IQR Q3 Q1 0.7 0.5 1.2
16. (a) Min 1.3
(b) Max 2.1
(c) Q1 0.3
(d) Q2 0.2
(e) Q3 0.4
(f) IQR Q3 Q1 0.4 0.3 0.7
17. None. The data are not skewed or symmetric. 18. Skewed right. Most of the data lie to the right. 19. Skewed left. Most of the data lie to the left. 20. Symmetric 21. Q1 B,
Q2 A,
Q3 C
25% of the entries are below B, 50% are below A, and 75% are below C. 22. P10 T,
P50 R,
P80 S
10% of the entries are below T, 50% are below R, and 80% are below S. 23. (a) Q1 2, (b)
Q2 4,
Q3 5
Watching Television
0
2
4 5
9
0 1 2 3 4 5 6 7 8 9
Hours
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CHAPTER 2
24. (a) Q1 2, (b)
Q2 4.5,
 DESCRIPTIVE STATISTICS
Q3 6.5
Vacation Days
0
2
4.5 6.5
0
2
4
6
10 8
10
Number of days
25. (a) Q1 3, (b)
Q2 3.85,
Q3 5.28
Airline Distances
1.8 3 3.85 5.28 6 0
1
2
3
4
5
6
Distances (in miles)
26. (a) Q1 15.125, Q2 15.8, Q3 17.65 (b)
Railroad Equipment Manufacturers
13.8 15.125
17.65 19.45 15.8
13.5 14.5 15.5 16.5 17.5 18.5 19.5
Hourly earnings (in dollars)
27. (a) 5
(b) 50%
(c) 25%
28. (a) $17.65
(b) 50%
(c) 50%
29. A ⇒ z 1.43 B⇒z0 C ⇒ z 2.14 The zscore 2.14 is unusual because it is so large. 30. A → z 1.54 B → z 0.77 C → z 1.54 None of the zscores are unusual. 31. (a) Statistics: x 73 ⇒ z
x 73 63 1.43 7
Biology: x 26 ⇒ z
x 26 23 0.77 3.9
(b) The student had a better score on the statistics test. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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32. (a) Statistics: x 60 ⇒ z Biology: x 20 ⇒ z
60 63 x 0.43 7
x 20 23 0.77 3.9
(b) The student had a better score on the statistics test. 33. (a) Statistics: x 78 ⇒ z
x 78 63 2.14 7
Biology: x 29 ⇒ z
x 29 23 1.54 3.9
(b) The student had a better score on the statistics test. 34. (a) Statistics: x 63 ⇒ z Biology: x 23 ⇒ z
x 63 63 0 7
x 23 23 0 3.9
(b) The student performed equally well on the two tests. 35. (a) x 34,000 ⇒ z
x 34,000 35,000 0.44 2,250
x 37,000 ⇒ z
x 37,000 35,000 0.89 2,250
x 31,000 ⇒ z
x 31,000 35,000 1.78 2,250
None of the selected tires have unusual life spans. (b) x 30,500 ⇒ z
x 30,500 35,000 2 ⇒ 2.5th percentile 2,250
x 37,250 ⇒ z
x 37,250 35,000 1 ⇒ 84th percentile 2,250
x 35,000 ⇒ z
x 35,000 35,000 0 ⇒ 50th percentile 2,250
36. (a) x 34 ⇒ z
x 34 33 0.25 4
x 30 ⇒ z
x 30 33 0.75 4
x 42 ⇒ z
x 42 33 2.25 4
The life span of 42 days is unusual due to a rather large zscore.
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CHAPTER 2
(b) x 29 ⇒ z
x 29 33 1 ⇒ 16th percentile 4
x 41 ⇒ z
x 41 33 2 ⇒ 97.5th percentile 4
x 25 ⇒ z
x 25 33 2 ⇒ 2.5th percentile 4
 DESCRIPTIVE STATISTICS
37. 68.5 inches 40% of the heights are below 68.5 inches. 38. 99th percentile 99% of the heights are below 76 inches. x 74 69.6 1.47 3.0
39. x 74: z x 62: z
x 62 69.6 2.53 3.0
x 80: z
x 80 69.6 3.47 3.0
The height of 62 inches is unusual due to a rather small zscore. The height of 80 inches is very unusual due to a rather large zscore. 40. x 70: z
x 70 69.6 0.13 3.0
x 66: z
x 66 69.6 1.20 3.0
x 68: z
x 68 69.6 0.53 3.0
None of the heights are unusual. 41. x 71.1: z
x 71.1 69.6 0.5 3.0
Approximately the 70th percentile. 42. x 66.3: z
x 66.3 69.6 1.1 3.0
Approximately the 12th percentile. 43. (a) 27 28 31 32 32 33 35 36 36 36 36 37 38 39 39 40 40 40 41 41 41 42 42 42 42 42 42 43 43 43 44 44 45 45 46 47 47 47 47 47 48 48 48 48 48 49 49 49 49 49 49 50 50 51 51 51 51 51 51 52 52 52 53 53 54 54 54 54 54 54 54 54 55 56 56 56 57 57 57 59 59 59 60 60 60 61 61 61 62 62 63 63 63 63 64 65 67 68 74 82 Q1
Q1 42,
Q2 49,
Q2
Q3
Q3 56
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59
60

CHAPTER 2
(b)
DESCRIPTIVE STATISTICS
Ages of Executives
27 25
42 49 56 35
45
55
82 65
75
85
Ages
(c) Half of the ages are between 42 and 56 years. (d) About 49 years old x 49.62 and Q2 49.00, because half of the executives are older and half are younger. (e) The age groups 20–29, 70–79, and 80–89 would all be considered unusual because they lie more than two standard deviations from the mean. 44. 1, 2, 3, 3, 5, 5, 7, 7, 8, 10 Q1
Q2
Midquartile 45. 22
23
24
Q3 Q1 Q3 3 7 5 2 2
32
33
34
Q1 28 Midquartile
36
38
39
40
41
47
Q3 39.5
Q2
Q1 Q3 28 39.5 33.75 2 2
46. 7.9, 8, 8.1, 9.7, 10.3, 11.2, 11.8, 12.2, 12.3, 12.7, 13.4, 15.4, 16.1 Q1 8.9 Midquartile
Q3 13.05
Q2
Q1 Q3 8.9 13.05 10.975 2 2
47. 13.4 15.2 15.6 16.7 17.2 18.7 19.7 19.8 19.8 20.8 21.4 22.9 28.7 30.1 31.9 Q1 Midquartile
Q2
Q3
Q1 Q3 16.7 22.9 19.8 2 2
48. (a) Disc 1: Symmetric Disc 2: Skewed left Disc 1 has less variation. (b) Disc 2 is more likely to have outliers because its distribution is wider. (c) Disc 1, because the distribution’s typical distance from the mean is roughly 16.3.
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CHAPTER 2
49.
Credit Card Purchases Friends: 75 102.5 136 159 190
You: 28 0
83 115 143
215
25 50 75 100 125 150 175 200 225
Monthly purchases (in dollars)
You
Friends
min 28
min 75
Q1 83
Q1 102.5
Q2 115
Q2 136
Q3 143
Q3 159
max 215
max 190
 DESCRIPTIVE STATISTICS
Your distribution is symmetric and your friend’s distribution is uniform. 50. Percentile 51. Percentile
Number of data values less than x Total number of data values 51 80
100
100 64th percentile
Number of data values less than x Total number of data values 75 80
100
100 94th percentile
CHAPTER 2 REVIEW EXERCISE SOLUTIONS 1. Class
Midpoint
Boundaries
Frequency, f
Relative frequency
Cumulative frequency
20–23 24–27 28–31 32–35 36–39
21.5 25.5 29.5 33.5 37.5
19.5–23.5 23.5–27.5 27.5–31.5 31.5–35.5 35.5–39.5
1 2 6 7 4
0.05 0.10 0.30 0.35 0.20
1 3 9 16 20
f 20
2.
f 1 n
37.5
33.5
29.5
25.5
21.5
Relative frequency
Income of Employees 0.35 0.30 0.25 0.20 0.15 0.10 0.05
Income (in thousands of dollars)
Greatest relative frequency: 32–35 Least relative frequency: 20–23
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61
3.
DESCRIPTIVE STATISTICS
4.
8 6 4
0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 11.875 11.915 11.955 11.995 12.035 12.075 12.115
12.115
12.075
12.035
11.995
11.955
11.915
2
11.875
Frequency
Liquid Volume 12oz Cans Relative frequency
Liquid Volume 12oz Cans
Actual volume (in ounces)
Actual volume (in ounces)
5. Class
Midpoint, x
Frequency, f
Cumulative frequency
79–93 94–108 109–123 124–138 139–153 154–168
86 101 116 131 146 161
9 12 5 3 2 1
9 21 26 29 31 32
Rooms Reserved 14 12 10 8 6 4 2
f 32
6.
71 86 101 116 131 146 161 176

CHAPTER 2
Frequency
62
Number of rooms
168.5
153.5
138.5
123.5
93.5
108.5
35 30 25 20 15 10 5
78.5
Cumulative frequency
Rooms Reserved
Number of rooms
7. 1 3 7 8 9 2 012333445557889 3 11234578 4 347 5 1 8.
Average Daily Highs
12
22
32
42
52
Temperature (in °F)
Height of Buildings Number of stories
9. 60 55 50 45 40 35 30 25 20
400 500 600 700 800
Height (in feet)
It appears as height increases, the number of stories increases.
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CHAPTER 2
10.
 DESCRIPTIVE STATISTICS
6 5 4 3 2 1
1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006
Unemployment rate
U.S. Unemployment Rate
American Kennel Club
12.
140 120 100 80 60 40 20
Boxer
Dachshund
Beagle
Breed
13. x 9.1
15.
16.
median 42
mode 7
mode 42
Midpoint, x
Frequency, f
xf
21.5 25.5 29.5 33.5 37.5
1 2 6 7 4
21.5 51.0 177.0 234.5 150.0
n 20
xf 634
f
xf
0 1 2 3 4 5 6
13 9 19 8 5 2 4
0 9 38 24 20 10 24
n 60
x f 125
17. x
Labrador Retriever 33.60%
Golden Retriever Yorkshire 11.65% German Terrier Shepherd 13.01% 11.92%
xf 634 31.7 n 20
x
x
Boxer 9.49%
14. x 40.6
median 8.5
x
American Kennel Club
Dachshund 9.76% Beagle 10.57% Labrador Retriever Yorkshire Terrier German Shepherd Golden Retriever
11.
Number registered (in thousands)
Year
xf 125 2.1 n 60 xw 780.15 720.15 860.15 910.15 870.15 800.25 w 0.15 0.15 0.15 0.15 0.15 0.25 82.1 82.1 1
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63
64
CHAPTER 2
18. x

DESCRIPTIVE STATISTICS
xw 960.20 850.20 910.20 860.40 w 0.20 0.20 0.20 0.40 88.8 88.8 1
19. Skewed
20. Skewed
21. Skewed left
22. Skewed right
23. Median
24. Mean
25. Range Max Min 8.26 5.46 2.8 26. Range Max Min 19.73 15.89 3.84 27.
x 96 6.9 14
x N 4 6.9 2
295.7 14
28.
2
21.12 4.6
x 602 66.9 N 9
x N
2
52 66.9
862.87 9
29. x s
2
s
86 66.92 . . . 68 66.92 56 66.92 9
95.87 9.8
x 36,801 2453.4 n 15
nx 1x 2445 2453.4 2
1,311,783.6 14
30. x
2 6.92 . . . 3 6.92 3 6.92 12
2
2.377 2453.42 14
93,698.8 306.1
x 416,659 52,082.4 n 8
x x2 n1
73,225,929.87 7
49,632 52,082.32 . . . 49,924 52,082.32 7
10,460,847.12 3234.3
31. 99.7% of the distribution lies within 3 standard deviations of the mean.
3 49 32.50 41.5 3 49 32.50 56.5 99.7% of the distribution lies between $41.50 and $56.50.
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CHAPTER 2
 DESCRIPTIVE STATISTICS
32. 46.75, 52.25 → 49.50 12.75, 49.50 12.75 → , 68% of the cable rates lie between $46.75 and $52.25.
36
33. n 40
8
20, 52 → 36 28, 36 28 ⇒ 2 , 2 ⇒ k 2 1
1 1 1 1 2 1 0.75 k2 2 4
At least 400.75 30 customers have a mean sale between $20 and $52.
7
34. n 20
2
3, 11 → 7 22, 7 22 → 2 , 2 → k 2 1
1 1 1 1 2 1 0.75 2 k 2 4
At least 200.75 15 shuttle flights lasted between 3 days and 11 days. 35. x
xf 99 2.5 n 40
nx 1x f 0 1.24 1 1 1.24398 2
s
2
2
. . . 5 1.2423
1.2 59.975 39
36. x s
xf 61 2.4 n 25
nx x1 f 2
0 2.44 4 1 2.44245
1.7 72.16 24
2
2
. . . 6 2.4421
37. Q1 56 inches 38. Q3 68 inches 39. IQR Q3 Q1 68 56 12 inches 40.
Height of Students
52 56 50
55
61 60
68 72 65
70
75
Heights
41. IQR Q3 Q1 33 29 4
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65
66 42.
CHAPTER 2

DESCRIPTIVE STATISTICS
Weight of Football Players
173 190 208
240
140 150 160 170 180 190 200 210 220 230 240
145
Weights
43. 23% of the students scored higher than 68. 44.
84 0.109 → 11% have larger audiences. 728 The station would represent the 89th percentile, P89.
45. x 213 ⇒ z
x 213 186 1.5 18
This player is not unusual. 46. x 141 ⇒ z
x 141 186 2.5 18
This is an unusually light player. 47. x 178 ⇒ z
x 178 186 0.44 18
This player is not unusual. 48. x 249 ⇒ z
x 249 186 3.5 18
This is an unusually heavy player.
CHAPTER 2 QUIZ SOLUTIONS
106.5 118.5 130.5 142.5 154.5
100.5–112.5 112.5–124.5 124.5–136.5 136.5–148.5 148.5–160.5
3 11 7 2 2
101–112 113–124 125–136 137–148 149–160
(b) Frequency Histogram and Polygon
Relative frequency 0.12 0.44 0.28 0.08 0.08
Weekly Exercise Relative frequency
166.5
154.5
142.5
130.5
118.5
94.5
106.5
Frequency
Minutes
3 14 21 23 25
(c) Relative Frequency Histogram
Weekly Exercise 10 8 6 4 2
Cumulative frequency
0.40 0.32 0.24 0.16 0.08
154.5
Frequency, f
142.5
Class boundaries
130.5
Midpoint
118.5
Class limits
106.5
1. (a)
Minutes
(d) Skewed
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CHAPTER 2
(f)
10
154.5
5 94.5
Minutes
15
142.5
100 110 120 130 140 150 160
20
130.5
157
118.5
101 117.5 123 131.5
Weekly Exercise 25
106.5
Weekly Exercise
Cumulative frequency
(e)
 DESCRIPTIVE STATISTICS
Minutes
2
Footwear 17.46% Equipment 26.70%
Clothing
Clothing 13.01%
U.S. Sporting Goods 40 32 24 16 8 Footwear
(b)
U.S. Sporting Goods Recreational transport 42.83%
Equipment
3. (a)
13.0 nx x1 f 4055.04 24
Recreational transport
s
xf 3130.5 125.2 n 25
Sales (in billions of dollars)
2. x
Sales area
x 751.6 n
4. (a) x
median 784.5 mode none The mean best describes a typical salary because there are no outliers. (b) range Max Min 575 s2 s
x x2 48,135.1 n1
nx 1x
2
219.4
5. x 2s 155,000 2
15,000 $125,000
x 2s 155,000 2
15,000 $185,000
95% of the new home prices fall between $125,000 and $185,000. 6. (a) x 200,000 (b) x 55,000
z z
x 200,000 155,000 3.0 ⇒ unusual price 15,000
x 55,000 155,000 6.67 ⇒ very unusual price 15,000
(c) x 175,000
z
x 175,000 155,000 1.33 ⇒ not unusual 15,000
(d) x 122,000
z
x 122,000 155,000 2.2 ⇒ unusual price 15,000
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67
68
CHAPTER 2

DESCRIPTIVE STATISTICS
7. (a) Q1 76
Q2 80
Q3 88
(b) IQR Q3 Q1 88 76 12 (c)
Wins for Each Team
61 60
76 80 88 70
80
97
90
100
Number of wins
CUMULATIVE REVIEW FOR CHAPTERS 1 AND 2 1. Systematic sampling 2. Simple Random Sampling. However, all U.S. adults may not have a telephone. 35%
12%
Entitlement Mentality
11%
Entitlement mentality
Stress
Stress
18%
Personal needs
Personal Needs
Family issues
24%
35 30 25 20 15 10 5 Personal illness
Family Issues
Absenteeism at U.S. Companies Percent of People
3. Personal Illness
Reason for absence
4. $42,500 is a parameter because it is describing the average salary of all 43 employees in a company. 5. 28% is a statistic because it is describing a proportion within a sample of 1000 adults. 6. (a) x 83,500, s $1500
80,500 86,500 83,500 ± 21500 ⇒ 2 standard deviations away from the mean. Approximately 95% of the electrical engineers will have salaries between 80,500, 86,500. (b) 40.95 38 7. Sample, because a survey of 1498 adults was taken. 8. Sample, because a study of 232,606 people was done. 9. Census, because all of the members of the Senate were included. 10. Experiment, because we want to study the effects of removing recess from schools. 11. Quantitative: Ratio 12. Qualitative: Nominal
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CHAPTER 2
13. min 0
 DESCRIPTIVE STATISTICS
Number of Tornados by State
Q1 2 0 2 12.5
Q2 12.5
0
Q3 39
20
39 40
136 60
80 100 120 140
Number of tornados
max 136 14. x
0.1585 0.1592 0.1584 21589 0.4091 88.9 0.15 0.15 0.15 0.15 0.40
15. (a) x 5.49 median 5.4 mode none Median, because the distribution is not symmetric. (b) Range 4.1 s2 2.34 s 1.53 The standard deviation of tail lengths of alligators is 1.53 feet. 16. (a) The number of deaths due to heart disease for women is decreasing. (b) The study was only conducted over the past 5 years and deaths may not decrease in the next year. 17. Class width
87 0 10.875 ⇒ 11 8
Class limits
Class boundaries
Class midpoint
Frequency, f
Relative Cumulative frequency frequency
0–10 11–21 22–32 33–43 44–54 55–65 66–76 77–87
0.5–10.5 10.5–21.5 21.5–32.5 32.5–43.5 43.5–54.5 54.5–65.5 65.5–76.5 76.5–87.5
5 16 27 38 49 60 71 82
8 8 1 1 1 4 0 4
0.296 0.296 0.037 0.037 0.037 0.148 0.000 0.148
n 27
n 1
8 16 17 18 19 23 23 27
f
18. Skewed right 19. Relative frequency
Detroit Redwings Points Scored 0.30 0.25 0.20 0.15 0.10 0.05
Least relative frequency: 66–76 Greatest relative frequency: 0–10 and 11–21
5 16 27 38 49 60 71 82
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69
CHAPTER
Probability
3 3.1 BASIC CONCEPTS OF PROBABILITY
3.1 Try It Yourself Solutions 1ab. (1)
(2) Agree
M
Disagree
F
M
c. (1) 6 outcomes
No opinion
F
M
Agree
F
M
Disagree
F
M
F
No opinion
M
F
(2) 9 outcomes
d. (1) Let A Agree, D Disagree, N No Opinion, M Male and F Female Sample space 再AM, AF, DM, DF, NM, NF冎 (2) Let A Agree, D Disagree, N No Opinion, R Republican, De Democrat, O Other Sample space 再AR, ADe, AO, DR, DDe, DO, NR, NDe, NO冎 2a. (1) 6 outcomes
(2) 1 outcome
b. (1) Not a simple event
(2) Simple event
3a. Manufacturer: 4 Size: 3 Color: 6 b. 共4兲共3兲共6兲 72 c. F C
L
C
W R B G T Y
W R B G T Y
M
W R B G T Y
H
G
W R B G T Y
C
M
W R B G T Y
L
M
L W R B G T Y
W R B G T Y
W R B G T Y
W R B G T Y
T
C
M
W R B G T Y
L W R B G T Y
W R B G T Y
4a. (1) 26 choices for each letter (2) 26, 25, 24, 23, 22, 21 choices (3) 22, 26, 26, 26, 26, 26 choices b. (1) 26
26 26 26 26 26 308,915,776
(2) 26
25 24 23 22 21 165,765,600
(3) 22 26
26 26 26 26 261,390,272 71
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72
CHAPTER 3

PROBABILITY
5a. (1) 52
(2) 52
(c) 52
b. (1) 1
(2) 13
(c) 52
c. (1) P(7 of diamonds) (2) P(diamond)
1 ⬇ 0.0192 52
13 0.25 52
(2) P(diamond, heart, club, or spade)
52 1 52
6a. Event the next claim processed is fraudulent (Freq 4) b. Total Frequency 100 c. P(fraudulent claim)
4 0.04 100
7a. Frequency 54 b. Total of the Frequencies 1000 c. P(age 15 to 24)
54 0.054 1000
8a. Event salmon successfully passing through a dam on the Columbia River. b. Estimated from the results of an experiment. c. Empirical probability 180 0.18 1000 180 820 0.82 b. P(not age 45 to 54) 1 1000 1000 820 c. or 0.82 1000
9a. P(age 45 to 54)
10a. 16 b. 再T1, T2, T3, T4, T5} c. P共tail and less than 6兲 11a. 10 10 b.
5 0.3125 16
10 10 10 10 10 10,000,000
1 10,000,000
3.1 EXERCISE SOLUTIONS 1. (a) Yes, the probability of an event occurring must be contained in the interval [0, 1] or [0%, 100%]. (b) Yes, the probability of an event occurring must be contained in the interval [0, 1] or [0%, 100%]. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 3

PROBABILITY
(c) No, the probability of an event occurring cannot be less than 0. (d) Yes, the probability of an event occurring must be contained in the interval [0, 1] or [0%, 100%]. (e) Yes, the probability of an event occurring must be contained in the interval [0, 1] or [0%, 100%]. (f) No, the probability of an event cannot be greater than 1. 2. It is impossible to have more than a 100% chance of rain. 3. The fundamental counting principle counts the number of ways that two or more events can occur in sequence. 4. The law of large numbers states that as an experiment is repeated over and over, the probabilities found in an experiment will approach the actual probabilities of the event. 5. 再A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z冎; 26 6. 再ΗΗΗ, ΗΗΤ, ΗΤΗ, ΗΤΤ, ΤΗΗ, ΤΗΤ, ΤΤΗ, ΤΤΤ冎; 8 H
H
T
H
H
T
T H
H
T
T
H
T
7.
T
再共A, 兲, 共B, 兲, 共AB, 兲, 共O, 兲, 共A, 兲, 共B, 兲, 共AB, 兲, 共O, 兲冎 where 共A, 兲 represents positive Rhfactor with A blood type and 共A, 兲 represents negative Rhfactor with A blood type; 8.
+
A
− +
B
− +
AB
− +
O
−
8. 再共1, 1兲, 共1, 2兲, 共1, 3兲, 共1, 4兲, 共1, 5兲, 共1, 6兲, 共2, 1兲, 共2, 2兲, 共2, 3兲, 共2, 4兲, 共2, 5兲, 共2, 6兲, 共3, 1兲, 共3, 2兲, 共3, 3兲, 共3, 4兲, 共3, 5兲, 共3, 6兲, 共4, 1兲, 共4, 2兲, 共4, 3兲, 共4, 4兲, 共4, 5兲, 共4, 6兲, 共5, 1兲, 共5, 2兲, 共5, 3兲, 共5, 4兲, 共5, 5兲, 共5, 6兲, 共6, 1兲, 共6, 2兲, 共6, 3兲, 共6, 4兲, 共6, 5兲, 共6, 6兲冎; 36
1
2
3
1 2 3 4 5 6
4
5
6
1 2 3 4 5 6
1 2 3 4 5 6 1 2 3 4 5 6
1 2 3 4 5 6 1 2 3 4 5 6
9. Simple event because it is an event that consists of a single outcome. 10. Not a simple event because it is an event that consists of more than a single outcome. Number less than 200 再1, 2, 3 . . . 199冎
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73
74
CHAPTER 3

PROBABILITY
11. Not a simple event because it is an event that consists of more than a single outcome. king 再king of hearts, king of spades, king of clubs, king of diamonds冎 12. Simple event because it is an event that consists of a single outcome. 13. 共9兲共15兲 135
14. 共3兲共6兲共4兲 72
15. 共9兲共10兲共10兲共5兲 4500
16. 共2兲共2兲共2兲共2兲共2兲共2兲 64
17. False. If you roll a sixsided die six times, the probability of rolling an even number at least once is approximately 0.9844. 18. False. You flip a fair coin nine times and it lands tails up each time. The probability it will land heads up on the tenth flip is 0.5. 19. False. A probability of less than 0.05 indicates an unusual event. 20. True 21. b
22. d
23. c
24. a
25. Empirical probability because company records were used to calculate the frequency of a washing machine breaking down. 26. Classical probability because each outcome is equally likely to occur. 27. P(less than 1000)
999 ⬇ 0.159 6296
28. P共greater than 1000兲
5296 ⬇ 0.841 6296
29. P(number divisible by 1000)
6 ⬇ 0.000953 6296
30. P共number not divisible by 1000兲
6290 ⬇ 0.999 6296
31. P共A兲
1 ⬇0.042 24
32. P共B兲
3 0.125 24
33. P共C兲
5 ⬇ 0.208 24
34. P共D兲
1 ⬇ 0.042 24
35. (a) 10 10 36. (a) 26
10 1000
9 10 10 5 117,000
(b)
1 0.001 1000
(c)
999 0.999 1000
(b)
1 117,000
(c)
116,999 117,000
37. 再共SSS兲, 共SSR兲, 共SRS兲, 共SRR兲, 共RSS兲, 共RSR兲, 共RRS兲, 共RRR兲冎 38. 再共RRR兲冎 39. 再共SSR兲, 共SRS兲, 共RSS兲冎 40. 再共SSR兲, 共SRS兲, 共SRR兲, 共RSS兲, 共RSR兲, 共RRS兲, 共RRR兲冎
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CHAPTER 3

PROBABILITY
41. Let S Sunny Day and R Rainy Day (a)
S SSSS R SSSR
R
S SSRS R SSRR
S
S SRSS R SRSR
R
S SRRS R SRRR
S
S RSSS R RSSR
R
S RSRS R RSRR
S
S RRSS R RRSR
R
S RRRS R RRRR
S R
S R R
42.
(b) 再共SSSS兲, 共SSSR兲, 共SSRS兲, 共SSRR兲, 共SRSS兲, 共SRSR兲
S S
共SRRS兲, 共SRRR兲, 共RSSS兲, 共RSSR兲, 共RSRS兲, 共RSRR兲, 共RRSS兲, 共RRSR兲, 共RRRS兲, 共RRRR兲冎 (c) 再共SSSR兲, 共SSRS兲, 共SRSS兲, 共RSSS兲冎
Not defective Supplier #1 Defective Not defective Supplier #2 Defective Not defective Supplier #3 Defective
43. P(voted in 2006 Gubernatorial election) 44. P共did not vote Democratic兲
4,092,652 ⬇ 0.500 8,182,876
61,159,368 ⬇ 0.535 114,413,842
45. P(between 21 and 24)
8.5 ⬇ 0.060 142.1
46. P共between 35 and 44兲
27.7 ⬇ 0.195 142.1
47. P(not between 18 and 20) 1
5.8 ⬇ 1 0.041 ⬇ 0.959 142.1
48. P共not between 25 and 34兲 1
21.7 ⬇ 1 0.153 ⬇ 0.847 142.1
49. P(Ph.D.)
8 ⬇ 0.090 89
51. P(master’s)
21 ⬇ 0.236 89
53. (a) P(pink)
2 0.5 4
50. P共Associate兲
52. P共Bachelor’s兲 (b) P(red)
54. P共same coloring as one of its parents兲 55. P(service industry)
18 ⬇ 0.202 89
1 0.25 4
33 ⬇ 0.371 89
(c) P(white)
1 0.25 4
8 0.5 16
113,409 ⬇ 0.785 144,428
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56. P共manufacturing industry兲
16,377 ⬇ 0.113 144,428
57. P(not in service industry) 1 P共service industry兲 1 0.785 0.215 58. P共not in agriculture, forestry, or fishing industry兲 1 P共agriculture, forestry, or fishing industry兲 1
2206 冢144,428 冣 ⬇ 0.985
59. (a) P(at least 21)
87 ⬇ 0.739 118
(b) P(between 40 and 50 inclusive) (c) P(older than 69)
32 ⬇ 0.269 118
1 0.008 118
60. (a) P共less than $21兲 0.25 (b) P共between $21 and $50兲 0.50 (c) P共$30 or more兲 0.50 61. The probability of choosing a tea drinker who does not have a college degree. 62. The probability of choosing a smoker whose mother did not smoke. 63. (a)
Sum
P 冇sum冈
Probability
2 3 4 5 6 7 8 9 10 11 12
1兾36 2兾36 3兾36 4兾36 5兾36 6兾36 5兾36 4兾36 3兾36 2兾36 1兾36
0.028 0.056 0.083 0.111 0.139 0.167 0.139 0.111 0.083 0.056 0.028
(b) Answers will vary. (c) The answers in part (a) and (b) will be similar. 64. No, the odds of winning a prize are 1:6. (One winning cap and 6 losing caps) Thus, the statement should read, “one in seven game pieces win a prize.” 65. (a) P共event will occur兲
4 ⬇ 0.444 9
(b) P共event will not occur兲
5 ⬇ 0.556 9
66. 13 : 39 1 : 3 67. 39 : 13 3 : 1
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CHAPTER 3

PROBABILITY
3.2 CONDITIONAL PROBABILITY AND THE MULTIPLICATION RULE
3.2 Try It Yourself Solutions 1a. (1) 30 and 102
(2) 11 and 50
b. (1) P(not have gene) 2a. (1) Yes
30 ⬇ 0.294 102
ⱍ
(2) P共not have gene normal IQ兲
11 0.22 50
(2) No
b. (1) Dependent 3a. (1) Independent
(2) Independent (2) Dependent
b. (1) Let A 再 swimming through 1st dam冎 B 再swimming through 2nd dam冎
ⱍ
P共A and B兲 P共A兲 P共B A兲 共0.85兲 共0.85兲 0.723 (2) Let A 再 selecting a heart冎 B 再selecting a second heart冎
ⱍ
P共A and B兲 P共A兲 P共B A兲 4a. (1) Find probability of the event
12 ⬇ 0.059 冢13 52 冣 冢 51 冣 (2) Find probability of the complement of the event
b. (1) P(3 knee surgeries successful) 共0.90兲 共0.90兲 共0.90兲 0.729 (2) P(at least one knee surgery successful) 1 P共none are successful兲 1 共0.10兲 共0.10兲 共0.10兲 0.999 5a. (1)(2) A 再is female冎; B 再works in health field冎
ⱍ
b. (1) P共A and B兲 P共A兲P共B A兲 共0.65兲共0.25兲
ⱍ
(2) P共A and B 兲 P共A兲共1 P共B A兲兲 共0.65兲共0.75兲 c. (1) 0.1625 (2) 0.4875
3.2 EXERCISE SOLUTIONS 1. Two events are independent if the occurrence of one of the events does not affect the probability of the occurrence of the other event.
ⱍ
ⱍ
If P共B A兲 P共B兲 or P共A B兲 P共A兲, then Events A and B are independent.
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2. (a) Roll a die twice. The outcome of the 2nd roll is independent of the outcome of the 1st roll. (b) Draw two cards (without replacement) from a standard 52card deck. The outcome of the 2nd card is dependent on the outcome of the 1st card.
ⱍ
3. False. If two events are independent, P共A B兲 P共A兲. 4. False. If events A and B are independent, then P共A and B兲 P共A兲 P共B兲. 5. These events are independent because the outcome of the 1st card drawn does not affect the outcome of the 2nd card drawn. 6. These events are dependent because returning a movie after its due date affects the outcome of receiving a late fee. 7. These events are dependent because the sum of the rolls depends on which numbers were rolled first and second. 8. These events are independent because the outcome of the 1st ball drawn does not affect the outcome of the 2nd ball drawn. 9. Events: depression, breathingrelated sleeping disorder Dependent. People with depression are more likely to have a breathingrelated sleeping disorder. 10. Events: stress, ulcers Independent. Stress only irritates already existing ulcers. 11. Events: memory loss, use of Aspartame Independent. The use of Aspartame does not cause memory loss. 12. Events: diabetes, obesity Dependent. Societies in which cases of diabetes are rare, obesity is rare. 13. Let A 再 have mutated BRCA gene冎 and B 再develop breast cancer冎. Thus 1 1 8 , and P共B A兲 . P共B兲 , P共A兲 8 600 10
ⱍ
ⱍ
(a) P共B A兲
8 0.8 10
ⱍ
(b) P共A and B兲 P共A兲 P共B A兲
1 冢600 冣 冢108 冣 0.0013
ⱍ
(c) Dependent because P共B A兲 P共B兲.
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 3

PROBABILITY
14. Let A 再drives pickup truck冎 and B 再drives a Ford冎. Thus, 1 3 2 P共A兲 , P共B兲 , and P共A B兲 . 6 10 9
ⱍ
ⱍ
(a) P共A B兲
2 ⬇ 0.222 9
ⱍ
(b) P共A and B兲 P共P兲P共A B兲
冢103 冣冢29冣 0.067
ⱍ
(c) Dependent because P共A B兲 P共A兲. 15. Let A 再own a computer冎 and B 再summer vacation this year冎. (a) P共B兲
45 0.308 146
(b) P共A兲
57 0.390 146
ⱍ
(c) P共B A兲
46 ⬇ 0.807 57
ⱍ
(d) P共A and B兲 P共A兲P共B A兲
57 46 冢146 冣冢57冣 0.315
ⱍ
(e) Dependent, because P共B A兲 P共B兲. The probability that the family takes a summer vacation depends on whether or not they own a computer. 16. Let A 再male冎 and B 再Nursing major冎. (a) P共B兲
ⱍ
795 0.225 3537
(c) P共B A兲
(b) P共A兲
1110 0.314 3537
95 0.086 1110
ⱍ
(d) P共A and B兲 P共A兲P共B A兲
95 0.027 冢1110 3537 冣冢 1110 冣
ⱍ
(e) Dependent because P共B A兲 P共B兲.
ⱍ
17. Let A 再pregnant冎 and B 再multiple births冎. Thus P共A兲 0.35 and P共B A兲 0.28.
ⱍ
(a) P共A and B兲 P共A兲 P共B A兲 共0.35兲 共0.28兲 0.098
ⱍ
ⱍ
(b) P共B A兲 1 P共B A兲 1 0.28 0.72 (c) It is not unusual because the probability of a pregnancy and multiple births is 0.098. 18. Let A 再has the opinion that race relations have improved冎 and B 再has the opinion that the rate of civil rights progress is too slow冎. Thus, P共A兲 0.6 and P共B A兲 0.4.
ⱍ
ⱍ
(a) P共A and B兲 P共A兲 P共B A兲 共0.6兲 共0.4兲 0.24
ⱍ
ⱍ
(b) P共B A兲 1 P共B A兲 1 0.4 0.6 (c) It is not unusual because the probability of an adult who thinks race relations have improved and says civil rights progress is too slow is 0.24.
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19. Let A 再 household in U.S. has a computer冎 and B 再has Internet access冎.
ⱍ
P共A and B兲 P共A兲P共B A兲 共0.62兲共0.88兲 0.546 20. Let A 再survives bypass surgery冎 and B 再heart damage will heal冎.
ⱍ
P共A and B兲 P共A兲P共B A兲 共0.60兲共0.50兲 0.30 21. Let A 再 1st person is lefthanded冎 and B 再2nd person is lefthanded冎.
ⱍ
(a) P共A and B兲 P共A兲 P共A B兲
120 ⬇ 0.014 冢1000 冣 冢119 999 冣
ⱍ
(b) P共A and B 兲 P共A 兲 P共B A 兲
880 ⬇ 0.774 冢1000 冣 冢879 999 冣
(c) P(at least one is lefthanded) 1 P共A and B 兲 1 0.774 0.226 22. Let A 再1st bulb fails冎 and B 再2nd bulb fails冎.
ⱍ
(a) P共A and B兲 P共A兲 P共B A兲
冢123 冣冢112 冣 ⬇ 0.045
ⱍ
(b) P共A and B 兲 P共A 兲 P共B A 兲
冢129 冣冢118 冣 ⬇ 0.545
(c) P共at least one bulb failed兲 1 P共none failed兲 1 P共A and B 兲 1 0.545 0.455 23. Let A 再 have one month’s income or more冎 and B 再man冎. (a) P共A兲
138 ⬇ 0.481 287
ⱍ
66 ⬇ 0.465 142
ⱍ
62 ⬇ 0.449 138
(b) P共A B兲 (c) P共B A兲
ⱍ
(d) Dependent because P共A 兲 ⬇ 0.519 0.465 ⬇ P共A B兲 Whether a person has at least one month’s income saved depends on whether or not the person is male.
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CHAPTER 3

PROBABILITY
24. Let A 再$100 or more冎 and B 再purebred冎. (a) P共A兲
ⱍ
50 ⬇ 0.556 90
(b) P共B A 兲
21 0.525 40
ⱍ
(c) P共A and B 兲 P共A兲 P共B A兲
15 ⬇ 0.167 冢50 冣 冢 90 50 冣
ⱍ
(d) Dependent because P共A兲 ⬇ 0.556 0.417 ⬇ P共A B 兲 The probability that the owner spent $100 or more on health care depends on whether or not the dog was a mixed breed. 25. (a) P共all five have AB兲 共0.03兲 共0.03兲 共0.03兲 共0.03兲 共0.03兲 0.0000000243 (b) P共none have AB兲 共0.97兲 共0.97兲 共0.97兲 共0.97兲 共0.97兲 ⬇ 0.859 (c) P共at least one has AB兲 1 P共none have AB兲 1 0.859 0.141 26. (a) P共all three have O兲 共0.38兲 共0.38兲 共0.38兲 ⬇ 0.055 (b) P共none have O兲 共0.62兲 共0.62兲 共0.62兲 ⬇ 0.238 (c) P共at least one has O兲 1 P共none have O兲 1 0.238 0.762 27. (a) P共first question correct兲 0.2 (b) P共first two questions correct兲 共0.2兲 共0.2兲 0.04 (c) P共first three questions correct兲 共0.2兲3 0.008 (d) P共none correct兲 共0.8兲3 0.512 (e) P共at least one correct兲 1 P共none correct兲 1 0.512 0.488 28. (a) P共none are defective兲 共0.995兲3 ⬇ 0.985 (b) P共at least one defective兲 1 P共none are defective兲 1 0.985 0.015 (c) P共all are defective兲 共0.005兲3 0.000000125 29. P (all three products came from the third factory) 30. (a) P共all share same birthday兲
ⱍ
ⱍ
24
23
109 108 ⬇ 0.011
1 1 ⬇ 0.00000751 冢365 365 冣冢 365 冣冢 365 冣
(b) P共none share same birthday兲 31. P共A B兲
25 110
364 363 0.992 冢365 365 冣冢 365 冣冢 365 冣
P共A兲 P共B A兲 P共A兲 P共B A兲 P共A 兲 P共B A 兲
ⱍ
ⱍ
冢23冣 冢15冣 0.133 0.444 2 1 1 1 0.133 0.167 冢3冣 冢5冣 冢3冣 冢2冣 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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ⱍ
32. P共A B兲

PROBABILITY
ⱍ
P共A兲 P共B A兲 P共A兲 P共B A兲 P共A 兲 P共B A 兲
ⱍ
ⱍ
冢38冣 冢23冣 0.25 0.4 3 2 5 3 0.25 0.375 冢8冣 冢3冣 冢8冣 冢5冣 ⱍ
33. P共A B兲
ⱍ
34. P共A B兲 35. P共A兲
P共A兲P共B ⱍ A兲 共A兲P共B ⱍ A兲 P共A 兲P共B ⱍ A 兲
共0.25兲共0.3兲 0.075 0.167 共0.25兲共0.3兲 共0.75兲共0.5兲 0.075 0.375
ⱍ
P共A兲P共A B兲 P共A兲P共A B兲 P共A 兲P共B A 兲
ⱍ
ⱍ
共0.62兲共0.41兲 0.254 0.797 共0.62兲共0.41兲 共0.38兲共0.17兲 0.254 0.065
1 0.005 200
ⱍ P共BⱍA 兲 0.05
P共B A兲 0.80
ⱍ
(a) P共A B兲
ⱍ
P共A兲 P共B A兲 P共A兲 P共B A兲 P共A 兲 P共B A 兲
ⱍ
ⱍ
共0.005兲 共0.8兲 0.004 ⬇ 0.074 共0.005兲 共0.8兲 共0.995兲 共0.05兲 0.004 0.04975
ⱍ
(b) P共A B 兲
ⱍ
P共A 兲 P共B A 兲 P共A 兲 P共B A 兲 P共A 兲 P共B A 兲
ⱍ
ⱍ
共0.995兲 共0.95兲 0.94525 ⬇ 0.999 共0.995兲 共0.95兲 共0.005兲 共0.2兲 0.94525 0.001
36. (a) P(different birthdays)
364 365
363
342
365 . . . 365 ⬇ 0.462
(b) P(at least two have same birthday) 1 P共different birthdays兲 1 0.462 0.538 (c) Yes, there were 2 birthdays on the 118th day. (d) Answers will vary. 37. Let A 再 flight departs on time冎 and B 再flight arrives on time冎.
ⱍ
P共A B 兲
P共A and B 兲 共0.83兲 ⬇ 0.954 P共B 兲 共0.87兲
38. Let A 再 flight departs on time冎 and B 再flight arrives on time冎.
ⱍ
P共A B 兲
P共A and B 兲 共0.83兲 ⬇ 0.933 P共B 兲 共0.87兲
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 3

PROBABILITY
3.3 THE ADDITION RULE
3.3 Try It Yourself Solutions 1a. (1) None of the statements are true. (2) None of the statements are true. (3) All of the statements are true. b. (1) A and B are not mutually exclusive. (2) A and B are not mutually exclusive. (3) A and B are mutually exclusive. 2a. (1) Mutually exclusive
(2) Not mutually exclusive
b. (1) Let A 再 6冎 and B 再odd冎. P共A兲
1 3 1 and P共B兲 6 6 2
(2) Let A 再 face card冎 and B 再heart冎. P共A兲
12 13 3 , P共B兲 , and P共A and B兲 52 52 52
c. (1) P共A or B兲 P共A兲 P共B兲
1 1 ⬇ 0.667 6 2
(2) P共A or B兲 P共A兲 P共B兲 P共A and B兲
12 13 3 ⬇ 0.423 52 52 52
3a. A 再sales between $0 and $24,999冎 B 再sales between $25,000 and $49,000冎 b. A and B cannot occur at the same time → A and B are mutually exclusive c. P共A兲
3 5 and P共B兲 36 36
d. P共A or B兲 P共A兲 P共B兲 4a. (1) Mutually exclusive
5 3 ⬇ 0.222 36 36
(2) Not mutually exclusive
b. (1) P共B or AB兲 P共B兲 P共AB兲
45 16 ⬇ 0.149 409 409
(2) P共O or Rh兲 P共O兲 P共Rh兲 P共O and Rh兲
184 344 156 ⬇ 0.910 409 409 409
5a. Let A 再 linebacker冎 and B 再quarterback冎. P共A or B兲 P共A兲 P共B兲
32 11 ⬇ 0.169 255 255
b. P(not a linebacker or quarterback) 1 P共A or B兲 ⬇ 1 0.169 ⬇ 0.831 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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3.3 EXERCISE SOLUTIONS 1. P共A and B兲 0 because A and B cannot occur at the same time. 2. (a) Toss coin once: A 再head冎 and B 再tail冎 (b) Draw one card: A 再ace冎 and B 再spade冎 3. True 4. False, two events being independent does not imply they are mutually exclusive. Example: 1 Toss a coin then roll a 6sided die. Let A 再head冎 and B 再6 on die冎. P共B A兲 6 P共B兲 1 implies A and B are independent events. However, P共A and B兲 12 implies A and B are not mutually exclusive.
ⱍ
5. False, P共A or B兲 P共A兲 P共B兲 P共A and B兲 6. True 7. Not mutually exclusive because the two events can occur at the same time. 8. Mutually exclusive because the two events cannot occur at the same time. 9. Not mutually exclusive because the two events can occur at the same time. The worker can be female and have a college degree. 10. Not mutually exclusive because the two events can occur at the same time. 11. Mutually exclusive because the two events cannot occur at the same time. The person cannot be in both age classes. 12. Not mutually exclusive because the two events can occur at the same time. 13. (a) No, it is possible for the events 再overtime冎 and 再temporary help冎 to occur at the same time. (b) P共OT or temp兲 P共OT兲 P共temp兲 P共OT and temp兲
18 9 5 ⬇ 0.423 52 52 52
14. (a) Not mutually exclusive because the two events can occur at the same time. (b) P共W or business兲 P共W兲 P共business兲 P共W and business兲
1800 860 425 ⬇ 0.639 3500 3500 3500
15. (a) Not mutually exclusive because the two events can occur at the same time. A carton can have a puncture and a smashed corner. (b) P(puncture or corner) P(puncture) P(corner) P(puncture and corner) 0.05 0.08 0.004 0.126 16. (a) Not mutually exclusive because the two events can occur at the same time. (b) P共does not have puncture or does not have smashed edge兲 P共does not have puncture兲 P共does not have smashed edge兲 P共does not have puncture and does not have smashed edge兲 0.96 0.93 0.8928 0.997
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CHAPTER 3

PROBABILITY
17. (a) P共diamond or 7兲 P共diamond兲 P共7兲 P共diamond and 7兲
13 4 1 ⬇ 0.308 52 52 52
(b) P共red or queen兲 P共red兲 P共queen兲 P共red and queen兲
26 4 2 ⬇ 0.538 52 52 52
(c) P共3 or face card兲 P共3兲 P共face card兲 P共3 and face card兲
4 12 0 ⬇ 0.308 52 52
18. (a) P共6 or greater than 4兲 P共6兲 P共greater than 4兲 P共6 and greater than 4兲
1 2 1 0.333 6 6 6
(b) P共less than 5 or odd兲 P共less than 5兲 P共odd兲 P共less than 5 and odd兲
4 3 2 0.833 6 6 6
(c) P共3 or even兲 P共3兲 P共even兲 P共3 and even兲
1 3 0 0.667 6 6
19. (a) P共under 5兲 0.068 (b) P共not 65兲 1 P共65兲 1 0.147 0.853 (c) P(between 18 and 34) P(between 18 and 24 or between 25 and 34) P(between 18 and 24) P(between 25 and 34) 0.097 0.132 0.229 20. (a) P共2兲 0.298 (b) P共2 or more兲 1 P共1兲 1 0.555 0.445 (c) P共between 2 and 5兲 0.298 0.076 0.047 0.014 0.435 21. (a) P共not completely satisfied兲 1
102 0.900 1017
(b) P共somewhat satisfied or completely satisfied兲 P共somewhat satisfied兲 P共completely dissatisfied兲 326 132 458 ⬇ 0.450 1017 1017 1017 171 22. (a) P共not as good as used to be兲 ⬇ 0.170 1005
(b) P共too much violence or tickets too costly兲 P共too much violence兲 P共tickets too costly兲 1
624 322 302 1 ⬇ 0.379 冢1005 1005 冣 1005
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PROBABILITY
23. A 再male冎; B 再nursing major冎 (a) P共A or B兲 P共A兲 P共B兲 P共A and B兲
1110 795 95 0.512 3537 3537 3537
(b) P共A or B兲 P共A兲 P共B兲 P共A and B兲
2427 2742 1727 0.973 3537 3537 3537
(c) P共A or B兲 0.512 (d) No mutally exclusive. A male can be a nursing major. 24. A 再left handed冎; B 再man冎 (a) P共A or B兲 P共A兲 P共B兲 P共A and B兲
113 475 50 0.538 1000 1000 1000
(b) P共A or B兲 P共A兲 P共B兲 P共A and B兲
887 525 462 0.950 1000 1000 1000
(c) P共A or B兲 P共A兲 P共B兲 P共A and B兲
113 525 63 0.575 1000 1000 1000
(d) P共A and B兲
425 0.425 1000
(e) Not mutually exclusive. A woman can be righthanded. 25. A 再frequently冎; B 再occasionally冎; C 再not at all冎; D 再male冎 (a) P共A or B兲
428 886 0.461 2850 2850
(b) P共D or C兲 P共D兲 P共C兲 P共D and C兲
1378 1536 741 0.762 2850 2850 2850
(c) P共D or A兲 P共D兲 P共A兲 P共D and A兲
428 221 1472 0.589 2850 2850 2850
(d) P共D or A兲 P共D兲 P共A兲 P共D and A兲
1378 2422 1171 0.922 2850 2850 2850
(e) Not mutually exclusive. A female can be frequently involved in charity work.
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CHAPTER 3
26. (a) P共contacts or glasses兲 (b) P共male or wears both兲 (c) P共female or neither兲

PROBABILITY
1268 253 0.475 3203 3203 1538 545 177 0.595 3203 3203 3203
1665 1137 681 0.662 3203 3203 3203
(d) P共male or does not wear glasses兲
1538 253 1137 64 456 0.752 3203 3203 3203
(e) The events are mutally exclusive because both cannot happen at the same time. 27. Answers will vary. Conclusion: If two events, 再A冎 and 再B冎, are independent, P共A and B兲 P共A兲 P共B兲. If two events are mutually exclusive, P共A and B兲 0. The only scenario when two events can be independent and mutually exclusive is if P共A兲 0 or P共B兲 0. 28. P共A or B or C 兲 P共A兲 P共B兲 P共C 兲 P共A and B兲 P共A and C兲 P共B and C兲 P共A and B and C兲 0.40 0.10 0.50 0.05 0.25 0.10 0.03 0.63 29. P共A or B or C兲 P共A兲 P共B兲 P共C兲 P共A and B兲 P共A and C 兲 P共B and C兲 P共A and B and C 兲 0.38 0.26 0.14 0.12 0.03 0.09 0.01 0.55
3.4 COUNTING PRINCIPLES
3.4 Try It Yourself Solutions 1a. n 6 teams 2a. 8P3
b. 6! 720
8! 8! 8 7 6 5 4 3 2 共8 3兲! 5! 54321
18
7 6 336
b. There are 336 possible ways that the subject can pick a first, second, and third activity. 3a. n 12, r 4 b.
12P4
12! 12! 12 11 10 9 11,880 共12 4兲! 8!
4a. n 20, n1 6, n2 9, n3 5 b.
n! 20! 77,597,520 n1! n2! n3! 6! 9! 5!
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PROBABILITY
5a. n 20, r 3 b.
20C3
1140
c. There are 1140 different possible threeperson committees that can be selected from 20 employees. 1 ⬇ 0.003 6a. 20P2 380 b. 380 6! 180 distinguishable permutations. 7a. 1 favorable outcome and 1! 2! 2! 1! b. P共Letter兲 8a.
15C5
3003
1 ⬇ 0.006 180 b.
9a. 共5C3兲 共7C0兲 10
54C5
3,162,510
1 10
b.
12C3
c. 0.0009 220
c.
10 ⬇ 0.045 220
3.4 EXERCISE SOLUTIONS 1. The number of ordered arrangements of n objects taken r at a time. An exmple of a permutation is the number of seating arrangements of you and three friends. 2. A selection of r of the n objects without regard to order. An example of a combination is the number of selections of different playoff teams from a volleyball tournament. 3. False, a permutation is an ordered arrangement of objects. 4. True
5. True
7. 7P3 8.
14P2
10. 8P6 11.
24C6
7! 7! 5040 210 共7 3兲! 4! 24 14! 14! 共14兲共13兲 . . . 共3兲共2兲共1兲 14 13 182 共14 2兲! 12! 共12兲共11兲 . . . 共3兲共2兲共1兲
9. 7C4
6. True
7! 共7兲共6兲共5兲共4兲共3兲共2兲共1兲 540 35 4!共7 4兲! 关共4兲共3兲共2兲共1兲兴关共3兲共2兲共1兲兴 共24兲共6兲 8! 40,320 8! 20,160 共8 6兲! 2! 2
24! 24! 134,596 6!共24 6兲! 6!18!
8! 8! C 4! 共 8 4 兲 ! 4!4! 70 0.076 12. 8 4 C 12! 12! 924 12 6 6!共12 6兲! 6!6! 6! 6! 720 共6 2兲! 4! 24 0.0060 13. P 10! 10! 3,628,800 10 4 共10 4兲! 6! 720 6P2
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CHAPTER 3
8! 8! C 3! 共 8 3 兲 ! 3! 5! 8! 14. 8 3 C 12! 12! 3! 5! 12 3 3!共12 3兲! 3! 9!

PROBABILITY
3! 9! 共8!兲共9!兲 12! 共5!兲共12!兲
关共18兲共7兲共6兲共5兲 . . . 共2兲共1兲兴关共9兲共8兲共7兲 . . . 共2兲共1兲兴 共8兲共7兲共6兲 0.255 关共5兲共4兲 . . . 共2兲共1兲兴关共12兲共11兲 . . . 共1兲兴 共12兲共11兲共10兲
15. Permutation, because order of the 15 people in line matters. 16. Combination, because order of the committee members does not matter. 17. Combinations, because the order of the captains does not matter. 18. Permutation, because the order of the letters matters. 19. 10 8
13C2 6240
21. 6! 720 20,358,520
23.
52C6
25.
18! 9,189,180 4! 8! 6!
20. 8! 40,320 22. 10! 3,628,800 24. 4! 24 26.
20C4
4845
27. 3S’s, 3T’s, 1A, 2I’s, 1C 10! 50,400 3! 3! 1! 2! 1! 28.
40C12
5,586,853,480
29. (a) 6! 720 (b) sample (c)
1 0.0014 720
31. (a) 12 (b) tree (c)
1 0.0833 12
33. (a) 907,200 (b) population (c)
1 0.000001 907,200
35.
1 1 0.0045 C 220 12 3
36.
1 1 0.012 84 9C3
30. (a) 60 (b) event (c)
1 0.017 60
32. (a) 720 (b) median (c)
1 0.0014 720
34. (a) 39,916,800 (b) distribution (c)
1 0.00000003 39,916,800
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CHAPTER 3

PROBABILITY
37. (a)
14 13 0.0164 冢15 56 冣冢 55 冣冢 54 冣
(b)
40 39 0.385 冢41 56 冣冢 55 冣冢 54 冣
38. (a)
冢146 冣冢135 冣冢124 冣冢113 冣 0.015
(b)
冢148 冣冢137 冣冢126 冣冢115 冣 0.070
39. (a) 8C4 70 (b) 2 2
冤
(c) 4C2
2 2 16
共2C0 兲 共2C0 兲 共2C2 兲 共2C2 兲 ⬇ 0.086 8C4
冥
40. (a) 26
26 10 10 10 10 6,760,000
(b) 24
24 10 10 10 10 5,760,000
(c) 0.50 41. (a) 共26兲共26兲共10兲共10兲共10兲共10兲共10兲 67,600,000 (b) 共26兲共25兲共10兲共9兲共8兲共7兲共6兲 19,656,000 (c)
1 ⬇ 0.000000015 67,600,000
42. (a) 共10兲共10兲共10兲 1000 (b) 共8兲共10兲共10兲 800 (c)
共8兲共10兲共5兲 1 0.5 共8兲共10兲共10兲 2
43. (a) 5! 120
(b) 2! 3! 12
(c) 3! 2! 12
(d) 0.4
44. (a) 共8C3兲 共2C0兲 共56兲 共1兲 56 (b) 共8C2兲 共2C1兲 共28兲 共2兲 56 (c) At least two good units one or fewer defective units. 56 56 112 (d) P共at least 2 defective units兲
共8C1兲 共2C2兲 共8兲 共1兲 ⬇ 0.067 120 10C3
45. 共8%兲共1200兲 共0.08兲共1200兲 96 of the 1200 rate financial shape as excellent. P共all four rate excellent兲
96C4 1200C4
3,321,960 ⬇ 0.000039 85,968,659,700
46. 共14%兲共1200兲 共0.14兲共1200兲 168 of 1200 rate financial shape as poor. P共all 10 rate poor兲
168C10 1200C10
⬇ 2.29 109
47. 共36%兲共500兲 共0.36兲共500兲 180 of the 1500 rate financial shape as fair ⇒ 500 180 320 rate shape as not fair. P共none of 80 selected rate fair兲
320C80 500C80
⬇ 5.03
1018
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CHAPTER 3

PROBABILITY
48. 共41%兲共500兲 共0.41兲共500兲 205 of 500 rate financial shape as good ⇒ 500 205 295 rate as not good. P共none of 55 selected rate good兲
295C55 500C55
⬇ 2.52
40C5
50. (a)
200C15
1.4629
1014
(b) P共win兲
658,008
49. (a)
1022
(b)
144C15
1 ⬇ 0.00000152 658,008
8.5323
1019
共144C15兲 ⬇ 0.00583 共200C15兲
(c) P共no minorities兲
(d) Yes, there is a very low probability of randomly selecting 15 nonminorities. 13C1 4C4 12C1 4C1
51. (a)
52C5 13C1 4C3 12C1 4C2
(b) (c)
52C5
共13兲共4兲共12兲共6兲 0.00144 2,598,960
52C5 13C2 13C1 13C1 13C1 52C5 24C8
52. (a)
53.
共13兲共1兲共12兲共4兲 0.0002 2,598,960
13C1 4C3 12C2 4C1 4C1
(d)
(c)
41C8
共13兲共4兲共66兲共4兲共4兲 0.0211 2,598,960
共78兲共13兲共13兲共13兲 0.0659 2,598,960
735,471 0.0077 95,548,245
24C4 17C2
14C4
41C8
(b)
共134,596兲共136兲 0.1916 95,548,245
(d)
17C8 41C8
24,310 0.0003 95,548,245
24C4 17C4 41C8
共10,626兲共2380兲 0.2647 95,548,245
1001 possible 4 digit arrangements if order is not important.
Assign 1000 of the 4 digit arrangements to the 13 teams since 1 arrangement is excluded. 54.
14P4
24,024
55. P共1st兲
250 0.250 1000
P共8th兲
28 0.028 1000 17 0.017 1000
P共2nd兲
199 0.199 1000
P共9th兲
P共3rd兲
156 0.156 1000
P共10th兲
11 0.011 1000
P共4th兲
119 0.119 1000
P共11th兲
8 0.008 1000
P共5th兲
88 0.088 1000
P共12th兲
7 0.007 1000
P共6th兲
63 0.063 1000
P共13th兲
6 0.006 1000
P共7th兲
43 0.043 1000
P共14th兲
5 0.005 1000
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91
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CHAPTER 3

PROBABILITY
56. Let A 再 team with the worst record wins second pick冎 and B 再team with the best record, ranked 14th, wins first pick冎.
ⱍ
P共A B兲
250 250 ⬇ 0.251 共1000 5兲 995
57. Let A 再team with the worst record wins third pick冎 and B 再team with the best record, ranked 13th, wins first pick冎 and C 再team ranked 2nd wins the second pick冎.
ⱍ
P共A B and C兲
250 0.314 796
58. Let A 再 neither the first nor the secondworst teams will get the first pick冎 and B 再the first or secondworst team will get the first pick冎. P共A兲 1 P共B兲 1
250 199 449 1冢 1 0.449 0.551 冢1000 冣 100 1000 冣
CHAPTER 3 REVIEW EXERCISE SOLUTIONS 1. Sample space: 再HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT冎 H H T H H T T
H H T T H T T
H T H T H T H T H T H T H T H T
Event: Getting three heads 再HHHT, HHTH, HTHH, THHH冎
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CHAPTER 3

PROBABILITY
2. Sample space: 再共1, 1兲, 共1, 2兲, 共1, 3兲, 共1, 4兲, 共1, 5兲, 共1, 6兲, 共2, 1兲, 共2, 2兲, 共2, 3兲, 共2, 4兲, 共2, 5兲, 共2, 6兲, 共3, 1兲, 共3, 2兲, 共3, 3兲, 共3, 4兲, 共3, 5兲, 共3, 6兲, 共4, 1兲, 共4, 2兲, 共4, 3兲, 共4, 4兲, 共4, 5兲, 共4, 6兲, 共5, 1兲, 共5, 2兲, 共5, 3兲, 共5, 4兲, 共5, 5兲, 共5, 6兲, 共6, 1兲, 共6, 2兲, 共6, 3兲, 共6, 4兲, 共6, 5兲, 共6, 6兲冎
1
1 2 3 4 5 6
2
1 2 3 4 5 6
3
4
1 2 3 4 5 6
5
1 2 3 4 5 6
6
1 2 3 4 5 6
1 2 3 4 5 6
Event: sum of 4 or 5 再共1, 3兲, 共1, 4兲, 共2, 2兲, 共2, 3兲, 共3, 1兲, 共3, 2兲, 共4, 1兲冎 3. Sample space: 再Jan, Feb . . ., Dec冎 Event: 再Jan, June, July冎 4. Sample space: 再GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB冎 G G B G B B
G GGG B GGB G GBG B GBB G BGG B BGB G BBG B BBB
Event: The family has two boys.
再BGB, BBG, GBB冎 5. 共7兲共4兲共3兲 84
6. 共26兲共26兲共26兲共10兲共10兲共10兲共10兲 175,760,000
7. Empirical probability
8. Classical probability
9. Subjective probability
10. Empirical probability
11. Classical probability
12. Empirical probability
13. P共at least 10兲 0.108 0.089 0.018 0.215 14. P共less than 20兲 0.608 .177 .108 0.893 15.
1 1.25 107 共8兲共10兲共10兲共10兲共10兲共10兲共10兲
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93
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CHAPTER 3
16. 1

PROBABILITY
1 0.999999875 共8兲共10兲共10兲共10兲共10兲共10兲共10兲
ⱍ
17. P共undergrad 兲 0.92
ⱍ
18. P共graduate 兲 0.07 19. Independent, the first event does not affect the outcome of the second event. 20. Dependent, the first event does affect the outcome of the second event. 21. P(correct toothpaste and correct dental rinse) P(correct toothpaste) P(correct dental rinse)
冢18冣 冢15冣 ⬇ 0.025
22. P共1st pair black and 2nd pair blue or white兲
ⱍ
P共1st pair black兲 P共2nd pair blue or white 1st black兲
0.2353 冢186 冣 冢12 17 冣
23. Mutually exclusive because both events cannot occur at the same time. 24. Not mutually exclusive because both events can occur at the same time. 25. P共home or work兲 P共home兲 P共work兲 P共home and work兲 0.44 0.37 0.21 0.60 26. P共silver or SUV兲 P共silver兲 P共SUV兲 P共silver and SUV兲 0.19 0.22 0.16 0.25 27. P共4–8 or club兲 P共4–8兲 P共club兲 P共4–8 and club兲
20 13 5 ⬇ 0.538 52 52 52
28. P共red or queen兲 P共red兲 P共queen兲 P共red and queen)
26 4 2 0.538 52 52 52
29. P共odd or less than 4兲 P共odd兲 P共less than 4兲 P共odd and less than 4兲
6 3 2 ⬇ 0.583 12 12 12
30. P共even or greater than 6兲 P共even兲 P共greater than 6兲 P共even and greater than 6兲
4 2 1 5 0.625 8 8 8 8
31. P共600 or more兲 P共600 999兲 P共1000 or more兲 0.198 0.300 0.498 32. P共300 999兲 P共300 599兲 P共600 999兲 0.225 0.198 0.423
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CHAPTER 3

PROBABILITY
33. P共poor taste or hard to find兲 P共 poor taste兲 P共hard to find兲 60 55 115 0.23 500 500 500
34. P共no time to cook or confused about nutrition兲 P共no time to cook兲 P共confused about nutrition兲
175 30 205 0.41 500 500 500
35. Order is important: 15P3 2730 36. 5! 120 37. Order is not important: 17C4 2380 38.
13C2
78
39. P共3 kings and 2 queens兲 40.
4C3
4C2
52C5
46 ⬇ 0.00000923 2,598,960
1 1 ⬇ 0.0000064 共23兲共26兲共26兲共10兲 155,480
41. (a) P共no defectives兲 (b) P共all defective兲
197C3 200C3 3C3
200C3
1,254,890 ⬇ 0.955 1,313,400
1 ⬇ 0.000000761 1,313,400
(c) P(at least one defective) 1 P(no defective) 1 0.955 0.045 (d) P(at least one nondefective) 1 P共all defective兲 1 0.000000761 ⬇ 0.999999239 42. (a) (b)
346 350
345
344
343
349 348 347 ⬇ 0.955
4 3 2 1 冢350 冣冢349 冣冢348 冣冢347 冣 0.000000002
(c) P共at least one winner兲 1 P共no winners兲 1 0.955 0.045 (d) P共at least one nonwinner兲 1 P共all winners兲 1
4 3 2 1 冢350 冣冢349 冣冢348 冣冢347 冣 0.999999998
CHAPTER 3 QUIZ SOLUTIONS 1399 ⬇ 0.524 2671 804 (b) P共bachelor F兲 ⬇ 0.515 1561 595 (c) P共bachelor M兲 ⬇ 0.536 1110
1. (a) P共bachelor兲
ⱍ
ⱍ
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95
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CHAPTER 3

PROBABILITY
(d) P共associate or bachelor兲 P共associate兲 P共bachelor兲
ⱍ
(e) P共doctorate M兲
665 1399 ⬇ 0.773 2671 2671
25 ⬇ 0.023 1110
(f) P共master or female兲 P共master兲 P共female兲 P共master and female兲
559 1561 329 0.671 2671 2671 2671
ⱍ
(g) P共associate and male兲 P共associate兲 P共male associate兲
ⱍ
(h) P共F bachelor兲
665 2671
260
665 ⬇ 0.097
804 ⬇ 0.575 1399
2. Not mutually exclusive because both events can occur at the same time. Dependent because one event can affect the occurrence of the second event. 3. (a)
518,665
147C3
(b) 3C 3 1 (c)
150C3
4. (a) (b) (c)
147C3 150C3 3C3 150C3
30P4
518,665 ⬇ 0.94 551,300
1 ⬇ 0.00000181 551,300
3C3 551,299 ⬇ 0.999998 551,300 150C3
150C3
5. 9 10 6.
3C3 551,300 1 551,299
10 10 10 5 450,000
657,720
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CHAPTER
Discrete Probability Distributions
4
4.1 PROBABILITY DISTRIBUTIONS 4.1 Try It Yourself Solutions 1a. (1) measured
(2) counted
b. (1) Random variable is continuous because x can be any amount of time needed to complete a test. (2) Random variable is discrete because x can be counted. x
f
P 冇c. x冈
0 1 2 3 4 5 6 7
16 19 15 21 9 10 8 2
0.16 0.19 0.15 0.21 0.09 0.10 0.08 0.02
n 100
New Employee Sales P(x)
Relative frequency
2ab.
0.20 0.15 0.10 0.05 x 0 1 2 3 4 5 6 7
Sales per day
兺
P共x兲 1
3a. Each P共x兲 is between 0 and 1. b. 兺P共x兲 1 c. Because both conditions are met, the distribution is a probability distribution. 4a. (1) Yes, each outcome is between 0 and 1. b. (1) Yes, 兺P冇x冈 1.
(2) Yes, 兺P共x兲 1.
c. (1) Is a probability distribution 5ab.
(2) Yes, each outcome is between 0 and 1.
x
P 冇x冈
xP 冇x冈
0 1 2 3 4 5 6 7
0.16 0.19 0.15 0.21 0.09 0.10 0.08 0.02
共0兲共0.16兲 0.00 共1兲共0.19兲 0.19 共2兲共0.15兲 0.30 共3兲共0.21兲 0.63 共4兲共0.09兲 0.36 共5兲共0.10兲 0.50 共6兲共0.08兲 0.48 共7兲共0.02兲 0.14
(2) Is a probability distribution
兺P共x兲 1 兺xP共x兲 2.60 c. 兺xP共x兲 2.6 On average, 2.6 sales are made per day.
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98
CHAPTER 4
6ab.

DISCRETE PROBABILITY DISTRIBUTIONS
x
P冇x冈
xⴚ
冇x ⴚ 冈2
P 冇x冈冇x ⴚ 冈2
0 1 2 3 4 5 6 7
0.16 0.19 0.15 0.21 0.09 0.10 0.08 0.02
2.6 1.6 0.6 0.4 1.4 2.4 3.4 4.4
6.76 2.56 0.36 0.16 1.96 5.76 11.56 19.36
共0.16兲共6.76兲 1.0816 共0.19兲共2.56兲 0.4864 共0.15兲共0.36兲 0.054 共0.21兲共0.16兲 0.0336 共0.09兲共1.96兲 0.1764 共0.10兲共5.76兲 0.576 共0.08兲共11.56兲 0.9248 共0.02兲共19.36兲 0.3872
兺P共x兲 1
兺P共x兲共x 兲2 3.72
c. 冪 2 冪3.720 ⬇ 1.9 d. A typical distance or deviation of the random variable from the mean is 1.9 sales per day. 7ab.
x
f
P 冇x冈
xP 冇x冈
0 1 2 3 4 5 6 7 8 9
25 48 60 45 20 10 8 5 3 1
0.111 0.213 0.267 0.200 0.089 0.044 0.036 0.022 0.013 0.004
共0兲共0.111兲 0.000 共1兲共0.213兲 0.213 共2兲共0.267兲 0.533 共3兲共0.200兲 0.600 共4兲共0.089兲 0.356 共5兲共0.044兲 0.222 共6兲共0.036兲 0.213 共7兲共0.022兲 0.156 共8兲共0.013兲 0.107 共9兲共0.004兲 0.040
n 225
兺
P共x兲 ⬇ 1
兺
xP共x兲 2.440
c. E共x兲 兺xP共x兲 3.08
Gain, x 1995 995 495 245 95 5
P 共x兲
xP 冇x冈
1 2000 1 2000 1 2000 1 2000 1 2000 1995 2000
1995 2000 995 2000 495 2000 245 2000 95 2000 9975 2000
兺P共x兲 1 兺xP共x兲 3.08
d. You can expect an average loss of $3.08 for each ticket purchased.
4.1 EXERCISE SOLUTIONS 1. A random variable represents a numerical value assigned to an outcome of a probability experiment. Examples: Answers will vary. 2. A discrete probability distribution lists each possible value a random variable can assume, together with its probability. Condition 1: 0 P共x兲 1 Condition 2: 兺P共x兲 1 3. An expected value of 0 represents the break even point, so the accountant will not gain or lose any money. 4. The mean of a probability distribution represents the “theoretical average” of a probability experiment. 5. False. In most applications, discrete random variables represent counted data, while continuous random variables represent measured data. 6. True 7. True 8. False. The expected value of a discrete random variable is equal to the mean of the random variable. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
9. Discrete, because home attendance is a random variable that is countable. 10. Continuous, because length of time is a random variable that has an infinite number of possible outcomes and cannot be counted. 11. Continuous, because annual vehiclemiles driven is a random variable that cannot be counted. 12. Discrete, because the number of fatalities is a random variable that is countable. 13. Discrete, because the number of motorcycle accidents is a random variable that is countable. 14. Continuous, because the random variable has an infinite number of possible outcomes and cannot be counted. 15. Continuous, because the random variable has an infinite number of possible outcomes and cannot be counted. 16. Discrete, because the random variable is countable. 17. Discrete, because the random variable is countable. 18. Continuous, because the random variable has an infinite number of possible outcomes and cannot be counted. 19. Continuous, because the random variable has an infinite number of possible outcomes and cannot be counted. 20. Discrete, because the random variable is countable. 21. (a) P共x > 2兲 0.25 0.10 0.35 (b) P共x < 4兲 1 P共4兲 1 0.10 0.90 22. (a) P共x > 1兲 1 P共x < 2兲 1 共0.30 0.25兲 0.45 (b) P共x < 3兲 0.30 0.25 0.25 0.80 23. 兺P共x兲 1 → P共3兲 0.22
24. 兺P共x兲 1 → P共1兲 0.15
25. Yes
26. No, 兺P共x兲 ⬇ 1.576
27. No, 兺P共x兲 0.95 and P共5兲 < 0.
28. Yes
29. (a)
x
f
P冇x冈
0 1 2 3 4 5
1491 425 168 48 29 14
0.686 0.195 0.077 0.022 0.013 0.006
n 2175
xP冇x冈
共0兲共0.686兲 0 共1兲共0.195兲 0.195 共2兲共0.077兲 0.155 共3兲共0.022兲 0.066 共4兲共0.013兲 0.053 共5兲共0.006兲 0.030
兺P共x兲 ⬇ 1 兺xP共x兲 0.497
冇x ⴚ 冈
冇x ⴚ 冈2
0.501 0.499 1.499 2.499 3.499 4.499
0.251 0.249 2.246 6.244 12.241 20.238
冇x ⴚ 冈2P冇x冈 共0.251兲共0.686兲 0.169 共0.249兲共0.195兲 0.049 共2.246兲共0.077兲 0.173 共6.244兲共0.022兲 0.138 共12.241兲共0.013兲 0.160 共20.238兲共0.006兲 0.122
兺共x 兲 P共x兲 0.811 2
(b) 兺xP共x兲 ⬇ 0.5 (c) 2 兺共x 兲2P共x兲 ⬇ 0.8 (d) 冪 2 ⬇ 冪0.8246 ⬇ 0.9 (e) A household on average has 0.5 dog with a standard deviation of 0.9 dog.
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99
100 30. (a)
CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
x
f
P冇x冈
xP冇x冈
冇x ⴚ 冈
冇 x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
0 1 2 3 4 5
1941 349 203 78 57 40
0.7275 0.1308 0.0761 0.0292 0.0214 0.0150
0 0.1308 0.1522 0.0876 0.0856 0.075
0.5321 0.4688 1.4688 2.4688 3.4688 4.4688
0.2822 0.2198 2.1574 6.0950 12.0326 19.9702
共0.2822兲共0.7275兲 0.2053 共0.2198兲共0.1308兲 0.0287 共2.1574兲共0.0761兲 0.1642 共6.0950兲共0.0292兲 0.1780 共12.0326兲共0.0214兲 0.2575 共19.9702兲共0.0150兲 0.2996
兺P共x兲 1 兺xP共x兲 0.531
n 2668
兺共x 兲 P共x兲 1.1333 2
(b) 兺xP共x兲 ⬇ 0.5 (c) 2 兺共x 兲2P共x兲 ⬇ 1.1 (d) 冪 2 ⬇ 1.1 (e) A household has on average 0.5 cat with a standard deviation of 1.1 cats. 31. (a)
x
f
P冇x冈
xP冇x冈
冇x ⴚ 冈
冇x ⴚ 冈2
0 1 2 3
300 280 95 20
0.432 0.403 0.137 0.029
0.000 0.403 0.274 0.087
0.764 0.236 1.236 2.236
0.584 0.056 1.528 5.000
n 695
兺P共x兲 ⬇ 1 兺xP共x兲 0.764
冇x ⴚ 冈2P冇x冈 共0.584兲共0.432兲 0.252 共0.056兲共0.403兲 0.022 共1.528兲共0.137兲 0.209 共5.000兲共0.029兲 0.145
兺共x 兲 P共x兲 0.629 2
(b) 兺xP共x兲 ⬇ 0.8 (c) 2 兺共x 兲2P共x兲 ⬇ 0.6 (d) 冪 2 ⬇ 0.8 (e) A household on average has 0.8 computer with a standard deviation of 0.8 computer. 32. (a)
x
f
P冇x冈
xP冇x冈
冇x ⴚ 冈
冇x ⴚ 冈2
0 1 2 3 4 5
93 113 87 64 13 8
0.250 0.297 0.229 0.168 0.034 0.021
0.000 0.297 0.458 0.505 0.137 0.105
1.503 0.503 0.497 1.497 2.497 3.497
2.259 0.253 0.247 2.241 6.235 12.229
n 380
兺P共x兲 1 兺xP共x兲 1.503
冇x ⴚ 冈2P冇x冈 共2.259兲共0.250兲 0.565 共0.253兲共0.297兲 0.752 共0.247兲共0.229兲 0.057 共2.241兲共0.168兲 0.377 共6.235兲共0.034兲 0.213 共12.229兲共0.021兲 0.257
兺共x 兲 P共x兲 1.545 2
(b) 兺xP共x兲 ⬇ 1.5 (c) 2 兺共x 兲2P共x兲 ⬇ 1.5 (d) 冪 2 ⬇ 1.2 (e) The average number of defects per batch is 1.5 with a standard deviation of 1.2 defects.
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CHAPTER 4
33. (a)

DISCRETE PROBABILITY DISTRIBUTIONS
x
f
P冇x冈
xP冇x冈
冇x ⴚ 冈
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
0 1 2 3 4 5 6
6 12 29 57 42 30 16
0.031 0.063 0.151 0.297 0.219 0.156 0.083
0.000 0.063 0.302 0.891 0.876 0.780 0.498
3.411 2.411 1.411 0.411 0.589 1.589 2.589
11.638 5.815 1.992 0.169 0.346 2.523 6.701
共11.638兲共0.031兲 0.360 共5.815兲共0.063兲 0.366 共1.992兲共0.151兲 0.300 共0.169兲共0.297兲 0.050 共0.346兲共0.219兲 0.076 共2.523兲共0.156兲 0.394 共6.701兲共0.083兲 0.557
n 192
兺P共x兲 1 兺xP共x兲 3.410
兺共x 兲 P共x兲 2.103 2
(b) 兺xP共x兲 ⬇ 3.4 (c) 2 兺共x 兲2P共x兲 ⬇ 2.1 (d) 冪 2 ⬇ 1.5 (e) An employee works an average of 3.4 overtime hours per week with a standard deviation of 1.5 hours. 34. (a)
x
f
P冇x冈
xP冇x冈
冇x ⴚ 冈
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
0 1 2 3 4 5 6 7
19 39 52 57 68 41 27 17
0.059 0.122 0.163 0.178 0.213 0.128 0.084 0.053
0.000 0.122 0.326 0.534 0.852 0.640 0.504 0.371
3.349 2.349 1.349 0.349 0.651 1.651 2.651 3.651
11.216 5.518 1.820 0.122 0.424 2.726 7.028 13.330
共11.216兲共0.059兲 0.662 共5.518兲共0.122兲 0.673 共1.820兲共0.163兲 0.297 共0.122兲共0.178兲 0.022 共0.424兲共0.213兲 0.090 共2.726兲共0.128兲 0.349 共7.028兲共0.084兲 0.590 共13.330兲共0.053兲 0.706
n 320
兺P共x兲 1
兺xP共x兲 3.349
兺共x 兲 P共x兲 3.389 2
(b) 兺xP共x兲 ⬇ 3.3 (c) 2 兺共x 兲2P共x兲 ⬇ 3.4 (d) 冪 2 ⬇ 1.8 (e) The average number of school related extracurricular activities per student is 3.3 with a standard deviation of 1.8 activities. 35.
x
P冇x冈
xP冇x冈
冇x ⴚ 冈
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
0 1 2 3 4 5 6 7 8
0.02 0.02 0.06 0.06 0.08 0.22 0.30 0.16 0.08
0.00 0.02 0.12 0.18 0.32 1.10 1.80 1.12 0.64
5.30 4.30 3.30 2.30 1.30 0.30 0.70 1.70 2.70
28.09 18.49 10.89 5.29 1.69 0.09 0.49 2.89 7.29
0.562 0.372 0.653 0.317 0.135 0.020 0.147 0.462 0.583
兺P共x兲 1 兺xP共x兲 5.30
兺共x 兲 P共x兲 3.250 2
(a) 兺xP共x兲 5.3
(b) 2 兺共x 兲2P共x兲 3.3
(c) 冪 2 1.8
(d) E关x兴 兺xP共x兲 5.3
(e) The expected number of correctly answered questions is 5.3 with a standard deviation of 1.8 questions.
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101
102 36.
CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
x
P冇x冈
xP冇x冈
冇x ⴚ 冈
0 1 2 3 4 5 6 7
0.01 0.10 0.26 0.32 0.18 0.06 0.03 0.03
0 0.10 0.52 0.96 0.72 0.30 0.18 0.21
2.99 1.99 0.99 0.01 1.01 2.01 3.01 4.01
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
8.94 3.96 0.98 0.0001 1.02 4.04 9.06 16.08
兺xP共x兲 2.99
0.089 0.396 0.254 0.000032 0.183 0.242 0.272 0.482
兺共x 兲 P共x兲 1.921 2
(a) 兺xP共x兲 3.0
(b) 2 兺共x 兲2P共x兲 1.9
(c) 冪 2 1.4
(d) E 关x兴 兺xP共x兲 3.0
(e) The expected number of 911 calls received per hour is 3.0 with a standard deviation of 1.4 calls. 37. (a) 兺xP共x兲 ⬇ 2.0 (c) 冪 2 ⬇ 1.0
(b) 2 兺共x 兲2P共x兲 ⬇ 1.0 (d) E 关x兴 兺xP共x兲 2.0
(e) The expected number of hurricanes that hit the U.S. is 2.0 with a standard deviation of 1.0. 38. (a) 兺xP共x兲 1.7 (c) 冪 2 ⬇ 1.0
(b) 2 兺共x 兲2P共x兲 ⬇ 1.0 (d) E关x兴 兺xP共x兲 1.7
(e) The expected car occupancy is 1.7 with a standard deviation of 1.0. 39. (a) 兺xP共x兲 ⬇ 2.5 (c) 冪 2 ⬇ 1.4
(b) 2 兺共x 兲2P共x兲 ⬇ 1.9 (d) E 关x兴 兺xP共x兲 2.5
(e) The expected household size is 2.5 persons with a standard deviation of 1.4 persons. 40. (a) 兺xP共x兲 1.6 (c) 冪 2 ⬇ 1.4
(b) 2 兺共x 兲2P共x兲 ⬇ 1.9 (d) E关x兴 兺xP共x兲 1.6
(e) The expected car occupancy is 1.6 with a standard deviation of 1.4. 41. (a) P共x < 2兲 0.686 0.195 0.881 (b) P共x 1兲 1 P共x 0兲 1 0.686 0.314 (c) P共1 x 3兲 0.195 0.077 0.022 0.294 42. (a) P共0兲 0.432 (b) P共x 1兲 1 P共0兲 1 0.432 0.568 (c) P共0 x 2兲 1 P共x 3兲 1 0.029 0.971 43. A household with three dogs is unusual because the probability is only 0.022. 44. A household with no computers is not unusual because the probability is 0.432. 45. E共x兲 兺xP共x兲 共1兲
1 共35兲 冢 冣 ⬇ $0.05 冢37 冣 38 38
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CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
46. E共x兲 兺xP共x兲 共3146兲
1 1 15 ⬇ $3.21 冢5000 冣 共446兲 冢5000 冣 共21兲 冢5000 冣 共4兲 冢4983 5000 冣
47. 4 a bx 1000 1.05共36,000兲 $38,800
ⱍⱍ
48. y b x 共1.04兲共3899兲 4054.96 49. xy x y 1532 1506 3038
xy x y 1532 1506 26 2 2 x2 y2 共312兲2 共304兲2 189,760 ⇒ xy 冪xy 435.61 50. xy
4.2 BINOMIAL DISTRIBUTIONS
4.2 Try It Yourself Solutions 1a. Trial: answering a question (10 trials) Success: question answered correctly b. Yes, the experiment satisfies the four conditions of a binomial experiment. c. It is a binomial experiment. n 10, p 0.25, q 0.75, x 0, 1, 2, . . . , 9, 10 2a. Trial: drawing a card with replacement (5 trials) Success: card drawn is a club Failure: card drawn is not a club b. n 5, p 0.25, q 0.75, x 3 c. P共3兲
5! 共0.25兲3共0.75兲2 ⬇ 0.088 2!3!
3a. Trial: selecting a worker and asking a question (7 trials) Success: Selecting a worker who will rely on pension Failure: Selecting a worker who will not rely on pension b. n 7, p 0.26, q 0.74, x 0, 1, 2, . . . , 6, 7 c. P共0兲 7C0共0.26兲0共0.74兲7 0.1215 P共1兲 7C1共0.26兲1共0.74兲6 0.2989 P共2兲 7C2共0.26兲2共0.74兲5 0.3150 P共3兲 7C3共0.26兲3共0.74兲4 0.1845 P共4兲 7C4共0.26兲4共0.74兲3 0.0648 P共5兲 7C5共0.26兲5共0.74兲2 0.0137 P共6兲 7C6共0.26兲6共0.74兲1 0.0016 P共7兲 7C7共0.26兲7共0.74兲0 0.0001
d.
x
P冇x冈
0 1 2 3 4 5 6 7
0.1215 0.2989 0.3150 0.1845 0.0648 0.0137 0.0016 0.0001
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103
104
CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
4a. n 250, p 0.71, x 178 b. P共178兲 ⬇ 0.056 c. The probability that exactly 178 people from a random sample of 250 people in the United States will use more than one topping on their hot dog is about 0.056. 5a. (1) x 2
(2) x 2, 3, 4, or 5
(3) x 0 or 1
b. (1) P共2兲 ⬇ 0.217 (2) P共0兲 5C0共0.21兲0共0.79兲5 0.308 P共1兲 5C1共0.21兲1共0.79兲4 0.409 P共x 2兲 1 P共0兲 P共1兲 1 0.308 0.409 0.283 or P共x 2兲 P共2兲 P共3兲 P共4兲 P共5兲 0.217 0.058 0.008 0.0004 0.283 (3) P共x < 2兲 P共0兲 P共1兲 0.308 0.409 0.717 c. (1) The probability that exactly two of the five men consider fishing their favorite leisuretime activity is about 0.217. (2) The probability that at least two of the five men consider fishing their favorite leisuretime activity is about 0.283. (3) The probability that fewer than two of the five men consider fishing their favorite leisuretime activity is about 0.717. 6a. Trial: selecting a business and asking if it has a Web site. (10 trials) Success: Selecting a business with a Web site Failure: Selecting a business without a site b. n 10, p 0.45, x 4 c. P共4兲 ⬇ 0.238 d. The probability of randomly selecting 10 small businesses and finding exactly 4 that have a website is 0.238. 7a. P共0兲 6C0共0.62兲0共0.38兲6 1共0.62兲0共0.38兲6 0.003 P共1兲 6C1共0.62兲1共0.38兲5 6共0.62兲1共0.38兲5 0.029 P共2兲 6C2共0.62兲0共0.38兲4 15共0.62兲2共0.38兲4 0.120 P共3兲 6C3共0.62兲3共0.38兲3 20共0.62兲3共0.38兲3 0.262 P共4兲 6C4共0.62兲4共0.38兲2 15共0.62兲4共0.38兲2 0.320 P共5兲 6C5共0.62兲5共0.38兲1 6共0.62兲5共0.38兲1 0.209 P共6兲 6C6共0.62兲6共0.38兲0 1共0.62兲6共0.38兲0 0.057
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CHAPTER 4
x
P冇x冈
0 1 2 3 4 5 6
0.003 0.029 0.120 0.262 0.320 0.209 0.057
DISCRETE PROBABILITY DISTRIBUTIONS
Owning a Computer
c.
P(x)
Relative frequency
b.

0.35 0.30 0.25 0.20 0.15 0.10 0.05 x 0 1 2 3 4 5 6
Households
8a. Success: Selecting a clear day n 31, p 0.44, q 0.56 b. np 共31兲共0.44兲 13.6 c. 2 npq 共31兲共0.44兲共0.56兲 7.6 d. 冪 2 2.8 e. On average, there are about 14 clear days during the month of May. The standard deviation is about 3 days.
4.2 EXERCISE SOLUTIONS 1. (a) p 0.50 (graph is symmetric) (b) p 0.20 (graph is skewed right → p < 0.5兲 (c) p 0.80 (graph is skewed left → p > 0.5兲 2. (a) p 0.75 共graph is skewed left → p > 0.5兲 (b) p 0.50 共graph is symmetric兲 (c) p 0.25 共graph is skewed right → p < 0.5兲 3. (a) n 12, 共x 0, 1, 2, . . . , 12兲 (b) n 4, 共x 0, 1, 2, 3, 4兲 (c) n 8, 共x 0, 1, 2, . . . , 8兲 As n increases, the probability distribution becomes more symmetric. 4. (a) n 10, 共x 0, 1, 2, . . . , 10兲 (b) n 15, 共x 0, 1, 2, . . . , 15兲 (c) n 5, 共x 0, 1, 2, 3, 4, 5兲 As n increases the probability distribution becomes more symmetric. 5. (a) 0, 1
(b) 0, 5
(c) 4, 5
6. (a) 0, 1, 2, 3, 4
(b) 0, 1, 2, 3, 4, 5, 6, 7, 8, 15
(c) 0, 1
7. Is a binomial experiment. Success: baby recovers n 5, p 0.80, q 0.20, x 0, 1, 2, . . . , 5 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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8. Is a binomial experiment. Success: person does not make a purchase n 18, p 0.74, q 0.26, x 0, 1, 2, . . . , 18 9. Is a binomial experiment. Success: Person said taxcuts hurt the economy. n 15, p 0.21, q 0.79; x 0, 1, 2, . . . , 15 10. Not a binomial experiment because the probability of a success is not the same for each trial. 11. np 共80兲共0.3兲 24
12. np 共64兲共0.85兲 54.4
2 npq 共80兲共0.3兲共0.7兲 16.8
2 npq 共64兲共0.85兲共0.15兲 ⬇ 8.2
冪 2 4.1
冪 2 ⬇ 2.9
13. np 共124兲共0.26兲 32.24
14. np 共316兲共0.72兲 ⬇ 227.5
2 npq 共124兲共0.26兲共0.74兲 23.858
2 npq 共316兲共0.72兲共0.28兲 ⬇ 63.7
冪 2 4.884
冪 2 ⬇ 8.0
15. n 5, p .25 (a) P共3兲 ⬇ 0.088 (b) P共x 3兲 P共3兲 P共4兲 P共5兲 0.088 0.015 .001 0.104 (c) P共x < 3兲 1 P共x 3兲 1 0.104 0.896 16. n 7, p 0.70 (a) P共5兲 0.318 (b) P共x 5兲 P共5兲 P共6兲 P共7兲 0.318 0.247 0.082 0.647 (c) P共x < 5兲 1 P共x 5兲 1 0.647 0.353 17. n 10, p 0.59 (using binomial formula) (a) P共8兲 0.111 (b) P共x 8兲 P共8兲 P共9兲 P共10兲 0.111 0.036 0.005 0.152 (c) P共x < 8兲 1 P共x 8兲 1 0.152 0.848 18. n 12, p 0.10 (a) P共4兲 ⬇ 0.021 (b) P共x 4兲 1 P共x < 4兲 1 P共0兲 P共1兲 P共2兲 P共3兲 1 0.282 0.377 0.230 0.085 0.026 (c) P共x < 4兲 1 P共x 4兲 1 0.026 0.974
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CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
19. n 10, p 0.21 (using binomial formula) (a) P共3兲 ⬇ 0.213 (b) P共x > 3兲 1 P共0兲 P共1兲 P共2兲 P共3兲 1 0.095 0.252 3.01 0.213 0.139 (c) P共x 3兲 1 P共x > 3兲 1 0.139 0.861 20. n 20, p 0.70 (use binomial formula) (a) P共1兲 ⬇ 1.627
109
(b) P共x > 1兲 P共0兲 P共1兲 ⬇ 1 3.487
1011 1.627
109 0.9999999983 ⬇ 1
(c) P共x 1兲 1 P共x > 1兲 ⬇ 1 0.9999999983 ⬇ 1.662 109 21. (a) P共3兲 0.028 (b) P共x 4兲 1 P共x 3) 1 共P共0兲 P共1兲 P共2兲 P共3兲兲 1 共0.000 0.001 0.007 0.028兲 0.964 (c) P共x 2兲 P共0兲 P共1兲 P共2兲 0.000 0.001 0.007 0.008 22. (a) P共2兲 0.264 (b) P共x > 6兲 P共7兲 P共8兲 P共9兲 P共10兲 共0.000 0.000 0.000 0.000兲 0.000 (c) P共x 5兲 P共0兲 P共1兲 P共2兲 P共3兲 0.221 0.360 0.264 0.115 0.033 0.006 0.999 23. (a) P共2兲 0.255 (b) P共x > 2兲 1 P共x 2兲1 共P共0兲 P共1兲 P共2兲兲 1 共0.037 0.146 0.255兲 0.562 (c) P共2 x 5兲 P共2兲 P共3兲 P共4兲 P共5兲 0.255 0.264 0.180 0.084 0.783
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DISCRETE PROBABILITY DISTRIBUTIONS
24. (a) P共4兲 0.183 (b) P共x > 4兲 1 P共x 4兲 1 共P共0兲 P共1兲 P共2兲 P共3兲 P共4兲兲 1 共0.037 0.141 0.244 0.257 0.183兲 0.138 (c) P共4 x 8兲 P共4兲 P共5兲 P共6兲 P共7兲 P共8兲 0.182 0.092 0.034 0.009 0.002 0.320
x
P冇x冈
0 1 2 3 4 5 6
0.063 0.220 0.323 0.253 0.112 0.026 0.003
Women Baseball Fans
(b)
P(x)
Probability
25. (a) n 6, p 0.37
(c) Skewed right
0.35 0.30 0.25 0.20 0.15 0.10 0.05 x 0 1 2 3 4 5 6
Women baseball fans
(d) np 共6兲共0.37兲 2.2 (e) 2 npq 共6兲共0.37兲共0.63兲 ⬇ 1.4 (f) 冪 2 ⬇ 1.2 (g) On average, 2.2 out of 6 women would consider themselves basketball fans. The standard deviation is 1.2 women. x 0, 5, or 6 would be unusual due to their low probabilities. 26. (a) n 5, p 0.25
(b)
(c) Skewed right
No Trouble Sleeping at Night
x
P冇x冈
0 1 2 3 4 5
0.237 0.396 0.264 0.088 0.015 0.001
Probability
P(x) 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05
x 0
1
2
3
4
5
Adults
(d) np 共5兲共0.25兲 1.3 (e) 2 npq 共5兲共0.25兲共0.75兲 ⬇ 0.9 (f) 冪 2 ⬇ 1.0 (g) On average, 1.3 adults out of every 5 have no trouble sleeping at night. The standard deviation is 1.0 adults. Four or five would be uncommon due to their low probabilities.
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CHAPTER 4
x
P冇x冈
0 1 2 3 4
0.814506 0.171475 0.013537 0.000475 0.000006
Donating Blood
(b)
DISCRETE PROBABILITY DISTRIBUTIONS
(c) Skewed right
P(x)
Probability
27. (a) n 4, p 0.05

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 x 0
1
2
3
4
Adults
(d) np 共4兲共0.05兲 0.2 (e) 2 npq 共4兲共0.05兲共0.95兲 0.2 (f) 冪 2 ⬇ 0.4 (g) On average, 0.2 eligible adult out of every 4 give blood. The standard deviation is 0.4 adult. x 2, 3, or 4 would be uncommon due to their low probabilities.
x
P冇x冈
0 1 2 3 4 5
0.092 0.281 0.344 0.211 0.065 0.008
Blood Type
(b) P(x)
Probability
28. (a) n 5, p 0.38
0.35 0.30 0.25 0.20 0.15 0.10 0.05 x 0
1
2
3
4
5
Adults with O+ blood
(c) Skewed right
(d) np 共5兲共0.38兲 1.9
(e) 2 npq 共5兲共0.38兲共0.62兲 1.2
(f) 冪 2 ⬇ 1.1
(g) On average, 1.9 adults out of every 5 have O blood. The standard deviation is 1.1 adults. Five would be uncommon due to its low probabilities. 29. (a) n 6, p 0.29 x
P冇x冈
0 1 2 3 4 5 6
0.128 0.314 0.321 0.175 0.053 0.009 0.001
30. (a) n 5, p 0.48 x
P冇x冈
0 1 2 3 4 5
0.038 0.175 0.324 0.299 0.138 0.025
(b) P共2兲 0.321 (c) P共at least 5兲 P共5兲 P共6兲 0.009 0.001 0.010
(b) P共2兲 0.325 (c) P共less than 4兲 1 P共4兲 P共5兲 1 0.138 0.025 0.837
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31. np 共6兲共0.29兲 1.7
2 npq 共6兲共0.29兲共0.71兲 1.2 冪 2 1.1 If 6 drivers are randomly selected, on average 1.7 drivers will name talking on cell phones as the most annoying habit of other drivers. Five of the six or all of the six randomly selected drivers naming talking on cell phones as the most annoying habit of other drivers is rare because their probabilities are less than 0.05. 32. np 共5兲共0.48兲 2.4
2 npq 共5兲共0.48兲共0.52兲 1.2 冪 2 1.1 If 5 employees are randomly selected, on average 2.4 employees would claim lack of time was a barrier. It would be unusual for no employees or all 5 employees to claim lack of time was a barrier because their probabilities are less than 0.05.
冢 冣 冢163 冣 冢163 冣 冢161 冣
33. P共5, 2, 2, 1兲
10! 9 5!2!2!1! 16
34. P共5, 2, 2, 1兲
10! 5 5!2!2!1! 16
5
2
2
1
冢 冣 冢164 冣 冢161 冣 冢166 冣 5
2
2
1
⬇ 0.033 ⬇ 0.002
4.3 MORE DISCRETE PROBABILITY DISTRIBUTIONS
4.3 Try It Yourself Solutions 1a. P共1兲 共0.23兲共0.77兲0 0.23 P共2兲 共0.23兲共0.77兲1 0.177 P共3兲 共0.23兲共0.77兲2 0.136 b. P共x < 4兲 P共1兲 P共2兲 P共3兲 0.543 c. The probability that your first sale will occur before your fourth sales call is 0.543. 30共2.71828兲3 ⬇ 0.050 0! 32共2.71828兲3 P共2兲 ⬇ 0.224 2! 34共2.71828兲3 P共4兲 ⬇ 0.168 4!
2a. P共0兲
31共2.71828兲3 ⬇ 0.149 1! 33共2.71828兲3 P共3兲 ⬇ 0.224 3! P共1兲
b. P共0兲 P共1兲 P共2兲 P共3兲 P共4兲 ⬇ 0.050 0.149 0.224 0.224 0.168 ⬇ 0.815 c. 1 0.815 ⬇ 0.185 d. The probability that more than four accidents will occur in any given month at the intersection is 0.185.
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CHAPTER 4
3a.

DISCRETE PROBABILITY DISTRIBUTIONS
2000 0.10 20,000
b. 0.10, x 3 c. P共3兲 0.0002 d. The probability of finding three brown trout in any given cubic meter of the lake is 0.0002.
4.3 EXERCISE SOLUTIONS 1. P共2兲 共0.60兲共0.4兲1 0.24
2. P共1兲 共0.25兲共0.75兲0 0.25
3. P共6兲 共0.09兲共0.91兲5 0.056
4. P共5兲 共0.38兲共0.62兲4 0.056
5. P共3兲
共4兲3共e 4兲 0.195 3!
6. P共5兲
共6兲5共e6兲 0.161 5!
7. P共2兲
共1.5兲2共e1.5兲 0.251 2!
8. P共4兲
共8.1兲4共e8.1兲 0.054 4!
9. The binomial distribution counts the number of successes in n trials. The geometric counts the number of trials until the first success is obtained. 10. The binomial distribution counts the number of successes in n trials. The Poisson distribution counts the number of occurrences that take place within a given unit of time. 11. Geometric. You are interested in counting the number of trials until the first success. 12. Poisson. You are interested in counting the number of occurrences that takes place within a given unit of time. 13. Poisson. You are interested in counting the number of occurrences that take place within a given unit of space. 14. Binomial. You are interested in counting the number of successes out of n trials. 15. Binomial. You are interested in counting the number of successes out of n trials. 16. Geometric. You are interested in counting the number of trials until the first success. 17. p 0.19 (a) P共5兲 共0.19兲共0.81兲4 ⬇ 0.082 (b) P(sale on 1st, 2nd, or 3rd call) P共1兲 P共2兲 P共3兲 共0.19兲共0.81兲0 共0.19兲共0.81兲1 共0.19兲共0.81兲2 ⬇ 0.469 (c) P共x > 3兲 1 P共x 3兲 1 0.469 0.531 18. p 0.526 (a) P共2兲 共0.526兲共0.474兲1 ⬇ 0.249 (b) P共makes 1st or 2nd shot兲 P共1兲 P共2兲 共0.526)共0.474兲0 共0.526兲共0.474兲1 ⬇ 0.775 (c) 共Binomial: n 2, p 0.526兲 P共0兲
2! 共0.526兲0共0.474兲2 ⬇ 0.225 0!2!
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CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
19. p 0.002 (a) P共10兲 共0.002兲共0.998兲9 ⬇ 0.002 (b) P(1st, 2nd, or 3rd part is defective) P共1兲 P共2兲 P共3兲 共0.002兲共0.998兲0 共0.002兲共0.998兲1 共0.002兲共0.998兲2 ⬇ 0.006 (c) P共x > 10兲 1 P共x 10兲 1 关P共1兲 P共2兲 . . . P共10兲兴 1 关0.020兴 ⬇ 0.980 20. p 0.25 (a) P共4兲 共0.25兲共0.75兲3 ⬇ 0.105 (b) P共prize with 1st, 2nd, or 3rd purchase兲 P共1兲 P共2兲 P共3兲 共0.25兲共0.75兲0 共0.25兲共0.75兲1 共0.25兲共0.75兲2 ⬇ 0.578 (c) P共x > 4兲 1 P共x 4兲 1 关P共1兲 P共2兲 P共3兲 P共4兲兴 1 关0.684兴 ⬇ 0.316 21. 3 (a) P共5兲
35e3 ⬇ 0.101 5!
(b) P共x 5兲 1 共P共0兲 P共1兲 P共2兲 P共3兲 P共4兲兲 ⬇ 1 共0.050 0.149 0.224 0.224 0.168兲 0.185 (c) P共x > 5兲 1 共P共0兲 P共1兲 P共2兲 P共3兲 P共4兲 P共5兲兲 ⬇ 1 共0.050 0.149 0.224 0.224 0.168 0.101兲 0.084 22. 4 (a) P共3兲 0.195 (b) P共x 3兲 P共0兲 P共1兲 P共2兲 P共3兲 0.0183 0.0733 0.1465 0.1954 0.433 (c) P共x > 3兲 1 P共x 3兲 1 0.4335 0.567 23. 0.6 (a) P共1兲 0.329 (b) P共x 1兲 P共0兲 P共1兲 0.549 0.329 0.878 (c) P共x > 1兲 1 P共x 1兲 1 0.878 0.122 24. 8.7 (a) P共9兲
8.79e8.7 ⬇ 0.131 9!
(b) P共x 9兲 0.627 (c) P共x > 9兲 1 P共x 9兲 1 0.627 0.373
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CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
25. (a) n 6000, p 0.0004 P共4兲 (b)
6000! 共0.0004兲4共0.9996兲5996 ⬇ 0.1254235482 5996!4!
6000 2.4 warped glass items per 2500. 2500
P共4兲 0.1254084986 The results are approximately the same. 26. (a) P共0兲 (b) P共1兲 (c) P共2兲
2C0 13C3 15C3 2C1 13C2 15C3 2C2 13C1 15C3
共1兲共286兲 ⬇ 0.629 共455兲
共2兲共78兲 ⬇ 0.343 共455兲
共1兲共13兲 ⬇ 0.029 共455兲
27. p 0.001 (a)
1 1 1000 p 0.001
2
q 0.999 999,000 2 p 共0.001兲2
冪 2 ⬇ 999.5 On average you would have to play 1000 times until you won the lottery. The standard deviation is 999.5 times. (b) 1000 times Lose money. On average you would win $500 every 1000 times you play the lottery. So, the net gain would be $500. 28. p 0.005 (a)
1 1 200 p 0.005
2
q 0.995 39,800 p2 共0.005兲2
冪 2 ⬇ 199.5 (b) 200 On average 200 records will be examined before finding one that has been miscalculated. The standard deviation is 199.5 records. 29. 3.8 (a) 2 3.8
冪 2 ⬇ 1.9 The standard deviation is 1.9 strokes. (b) P共x > 72兲 1 P共x 72兲 1 0.695 0.305 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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30. 7.6 (a) 2 7.6
冪 2 ⬇ 2.8 The standard deviation is 2.8 inches. (b) P共X > 12兲 1 P共X 12兲 ⬇ 1 0.954 ⬇ 0.046
CHAPTER 4 REVIEW EXERCISE SOLUTIONS 1. Discrete
2. Continuous
3. Continuous
4. Discrete
5. No, 兺P共x兲 1.
6. Yes
7. Yes
8. No, P共5兲 > 1 and 兺P共x兲 1. 10. No, 兺P共x兲 1.
9. Yes 11. (a)
x
Frequency
P冇x冈
xP冇x冈
xⴚ
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
2 3 4 5 6 7 8 9 10 11
3 12 72 115 169 120 83 48 22 6
0.005 0.018 0.111 0.177 0.260 0.185 0.128 0.074 0.034 0.009
0.009 0.055 0.443 0.885 1.560 1.292 1.022 0.665 0.338 0.102
4.371 3.371 2.371 1.371 0.371 0.629 1.629 2.629 3.629 4.629
19.104 11.362 5.621 1.879 0.137 0.396 2.654 6.913 13.171 21.430
0.088 0.210 0.623 0.332 0.036 0.073 0.339 0.510 0.446 0.198
n 650
(b)
兺P共x兲 1 兺xP共x兲 6.371
2
(c) 兺xP共x兲 ⬇ 6.4
Pages per Section P(x)
Probability
兺共x 兲 P共x兲 2.855
2 兺共x 兲2P共x兲 ⬇ 2.9
0.28 0.24 0.20 0.16 0.12 0.08 0.04
冪 2 ⬇ 1.7
x 2 3 4 5 6 7 8 9 1011
Pages
12. (a)
x
Frequency
P冇x冈
xP冇x冈
xⴚ
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
0 1 2 3 4 5
29 62 33 12 3 1
0.207 0.443 0.236 0.086 0.021 0.007
0.000 0.443 0.471 0.257 0.086 0.036
1.293 0.293 0.707 1.707 2.707 3.707
1.672 0.086 0.500 2.914 7.328 13.742
0.346 0.038 0.118 0.250 0.157 0.098
n 140
兺P共x兲 1 兺xP共x兲 1.293
兺共x 兲 P共x兲 1.007 2
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CHAPTER 4
(b)
DISCRETE PROBABILITY DISTRIBUTIONS
(c) 兺xP共x兲 ⬇ 1.3
Hits Per Game P(x)
2 兺共x 兲2P共x兲 ⬇ 1.0
0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05
Probability

冪 2 ⬇ 1.0 x 0
1
2
3
4
5
Hits
13. (a)
x
Frequency
P冇x冈
xP冇x冈
xⴚ
冇x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
0 1 2 3 4 5 6
3 38 83 52 18 5 1
0.015 0.190 0.415 0.260 0.090 0.025 0.005
0.000 0.190 0.830 0.780 0.360 0.125 0.030
2.315 1.315 0.315 0.685 1.685 2.685 3.685
5.359 1.729 0.099 0.469 2.839 7.209 13.579
0.080 0.329 0.041 0.122 0.256 0.180 0.068
n 200
兺P共x兲 1 兺xP共x兲 2.315
P(x)
Probability
2
(c) 兺xP共x兲 ⬇ 2.3
Televisions per Household
(b)
兺共x 兲 P共x兲 1.076
2 兺共x 兲2P共x兲 ⬇ 1.1
0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05
冪 2 ⬇ 1.0
x 0 1 2 3 4 5 6
Televisions
14. (a)
x 15 30 60 90 120
Frequency
P冇x冈
xP冇x冈
xⴚ
冇 x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
76 445 30 3 12
0.134 0.786 0.053 0.005 0.021
2.014 23.587 3.180 0.477 2.544
16.802 1.802 28.198 58.198 88.198
282.311 3.248 795.120 3386.993 7778.866
37.908 2.553 42.144 17.952 164.923
n 566
(b)
兺P共x兲 1 兺xP共x兲 31.802
2
(c) 兺xP共x兲 ⬇ 31.8
Advertising Sales P(x)
Probability
兺共x 兲 P共x兲 265.480
2 兺共x 兲2P共x兲 ⬇ 265.5
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
冪 2 ⬇ 16.3
x 15 30 60 90 120
Seconds
15. E共x兲 兺xP共x兲 3.4 16. E共x兲 兺xP共x兲 2.5 17. Yes, n 12, p 0.24, q 0.76, x 0, 1, . . . , 12. 18. No, the experiment is not repeated for a fixed number of trials.
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19. n 8, p 0.25 (a) P共3兲 0.208 (b) P共x 3兲 1 P共x < 3兲 1 关P共0兲 P共1兲 P共2兲兴 ⬇ 1 关0.100 0.267 0.311兴 0.322 (c) P共x > 3兲 1 P共x 3兲 1 关P共0兲 P共1兲 P共2兲 P共3兲兴 1 关0.100 0.267 0.311 0.208兴 0.114 20. n 12, p 0.25 (a) P共2兲 0.232 (b) P共x 2兲 1 P共0兲 P共1兲 1 0.0317 0.1267 0.842 (c) P共x > 2兲 1 P共0兲 P共1兲 P共2兲 1 0.032 0.127 0.232 0.609 21. n 7, p 0.43 (use binomial formula) (a) P共3兲 0.294 (b) P共x 3兲 1 P共x < 3兲 1 关P共0兲 P共1兲 P共2兲兴 ⬇ 1 关0.0195 0.1032 0.2336兴 0.644 (c) P共x > 3兲 1 P共x 3兲 ⬇ 1 关P共0兲 P共1兲 P共2兲 P共3兲兴 ⬇ 1 共0.0195 0.1032 0.2336 0.2937兲 0.350 22. n 5, p 0.31 (a) P共2兲 0.316 (b) P共x 2兲 P共2兲 P共3兲 P共4兲 P共5兲 0.316 0.142 0.032 0.003 ⬇ 0.492 (c) P共x > 2兲 P共3兲 P共4兲 P共5兲 ⬇ 0.142 0.032 0.003 ⬇ 0.177 x
P冇x冈
0 1 2 3 4 5
0.007 0.059 0.201 0.342 0.291 0.099
(b)
(c) np 共5兲共0.63兲 ⬇ 3.2
Renting Movies P(x)
Probability
23. (a)
2 npq 共5兲共0.63兲共0.37兲 1.2
0.35 0.30 0.25 0.20 0.15 0.10 0.05
冪 2 ⬇ 1.1
x 0
1
2
3
4
5
Americans
x
P冇x冈
0 1 2 3 4 5 6
0.001 0.014 0.073 0.206 0.328 0.279 0.099
(c) np 共6兲共0.68兲 4.1
Vacation Destinations
(b)
P(x)
Probability
24. (a)
2 npq 共6兲共0.68兲共0.32兲 1.3
0.35 0.30 0.25 0.20 0.15 0.10 0.05
冪 2 ⬇ 1.1 x 0 1 2 3 4 5 6
Families
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CHAPTER 4
x
P冇x冈
0 1 2 3 4
0.130 0.346 0.346 0.154 0.026
(b)
DISCRETE PROBABILITY DISTRIBUTIONS
(c) np 共4兲共0.40兲 ⬇ 1.6
Diesel Engines P(x)
Probability
25. (a)

2 npq 共4兲共0.40兲共0.60兲 1.0
0.35 0.30 0.25 0.20 0.15 0.10 0.05
冪 2 ⬇ 1.0
x 0
1
2
3
4
Trucks sold
26. n 5, p 0.15 x
P冇x冈
0 1 2 3 4 5
0.444 0.392 0.138 0.024 0.002 0.0001
(c) np 共5兲共0.15兲 0.8
Online Weather
(b)
P(x)
Probability
(a)
2 npq 共5兲共0.15兲共0.85兲 0.6
0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05
冪 2 ⬇ 0.8 x 0
1
2
3
4
5
People in U.S.
27. p 0.167 (a) P共4兲 ⬇ 0.096 (b) P共x 4兲 P共1兲 P共2兲 P共3兲 P共4兲 ⬇ 0.518 (c) P共x > 3兲 1 P共x 3兲 1 关P共1兲 P共2兲 P共3兲兴 ⬇ 0.579 28. p
73 0.477 153
(a) P共1兲 ⬇ 0.477
(b) P共2兲 ⬇ 0.249
(c) P共1 or 2兲 P共1兲 P共2兲 ⬇ 0.477 0.249 0.726 (d) P共X 3兲 P共1兲 P共2兲 P共3兲 0.477 0.249 0.130 0.856 29.
2457 68.25 ⬇ 68.25 lightning deaths兾year → ⬇ 0.1869 deaths兾day 36 365
(a) P共0兲
0.18690e 0.1869 ⬇ 0.830 0!
(b) P共1兲
0.18691e1869 ⬇ 0.155 1!
(c) P共x > 1兲 1 关P共0兲 P共1兲兴 1 关0.830 0.155兴 0.015 30. (a) 10 P共x 3兲 1 P共x < 3兲 1 关P共0兲 P共1兲 P共2兲兴 1 关0.0000 0.0005 0.0023兴 ⬇ 0.997 (b) 5 P共x 3兲 1 P共x < 3兲 1 关P共0兲 P共1兲 P共2兲兴 1 关0.0067 0.0337 0.0842兴 ⬇ 0.875 (c) 15 P共x 3兲 1 P共x < 3兲 1 关P共0兲 P共1兲 P共2兲兴 ⬇ 1 关0.0000 0.0000 0.0000兴 1 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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CHAPTER 4

DISCRETE PROBABILITY DISTRIBUTIONS
CHAPTER 4 QUIZ SOLUTIONS 1. (a) Discrete because the random variable is countable. (b) Continuous because the random variable has an infinite number of possible outcomes and cannot be counted. 2. (a)
x
Frequency
P冇x冈
xP冇x冈
xⴚ
冇 x ⴚ 冈2
冇x ⴚ 冈2P冇x冈
1 2 3 4 5
70 41 49 13 3
0.398 0.233 0.278 0.074 0.017
0.398 0.466 0.835 0.295 0.085
1.08 0.08 0.92 1.92 2.92
1.166 0.006 0.846 3.686 8.526
0.464 0.001 0.236 0.272 0.145
兺P共x兲 1 兺xP共x兲 2.080
n 176
(b)
兺共x 兲 P共x兲 1.119 2
Hurricane Intensity
Probability
P(x) 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 x 1
2
3
4
5
Intensity
(c) 兺xP共x兲 ⬇ 2.1
2 兺共x 兲2P共x兲 ⬇ 1.1 冪 2 ⬇ 1.1 On average the intensity of a hurricane will be 2.1. The standard deviation is 1.1. (d) P共x 4兲 P共4兲 P共5兲 0.074 0.017 0.091 3. n 8, p 0.80 x
P冇x冈
0 1 2 3 4 5 6 7 8
0.000003 0.000082 0.001147 0.009175 0.045875 0.146801 0.293601 0.335544 0.167772
(d) P共2兲 0.001
(c) np 共8兲共0.80兲 6.4
Successful Surgeries
(b)
P(x)
Probability
(a)
2 npq 共8兲共0.80兲共0.20兲 1.3
0.35 0.30 0.25 0.20 0.15 0.10 0.05
冪 2 ⬇ 1.1
x 0 1 2 3 4 5 6 7 8
Patients
(e) P共x < 2兲 P共0兲 P共1兲 0.000003 0.000082 0.000085
4. 5 (a) P共5兲 0.176 (b) P共x < 5兲 P共0兲 P共1兲 P共2兲 P共3兲 P共4兲 0.00674 0.03369 0.08422 0.14037 0.17547 ⬇ 0.440 (c) P共0兲 ⬇ 0.007
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CHAPTER
Normal Probability Distributions
5
5.1 INTRODUCTION TO NORMAL DISTRIBUTIONS AND THE STANDARD NORMAL DISTRIBUTION
5.1 Try It Yourself Solutions 1a. A: 45, B: 60, C: 45 (B has the greatest mean.) b. Curve C is more spread out, so curve C has the greatest standard deviation. 2a. Mean 3.5 feet b. Inflection points: 3.3 and 3.7 Standard deviation 0.2 foot 3a. (1) 0.0143
(2) 0.9850
4a.
5a.
z 0
−2.16
2.13
b. 0.9834
z 0
b. 0.0154 c. Area 1 0.0154 0.9846
6a. 0.0885
b. 0.0154
c. Area 0.0885 0.0154 0.0731
5.1 EXERCISE SOLUTIONS 1. Answers will vary.
2. 1
3. Answers will vary. Similarities: Both curves will have the same line of symmetry. Differences: One curve will be more spread out than the other. 4. Answers will vary. Similarities: Both curves will have the same shape (i.e., equal standard deviations) Differences: The two curves will have different lines of symmetry. 5. 0, 1 6. Transform each data value x into a zscore. This is done by subtracting the mean from x and dividing by the standard deviation. In symbols, z
x .
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
7. “The” standard normal distribution is used to describe one specific normal distribution 共 0, 1兲. “A” normal distribution is used to describe a normal distribution with any mean and standard deviation. 8. (c) is true because a zscore equal to zero indicates that the corresponding xvalue is equal to the mean. (a) and (b) are not true because it is possible to have a zscore equal to zero and the mean is not zero or the corresponding xvalue is not zero. 9. No, the graph crosses the xaxis. 10. No, the graph is not symmetric. 11. Yes, the graph fulfills the properties of the normal distribution. 12. No, the graph is skewed left. 13. No, the graph is skewed to the right. 14. No, the graph is not bellshaped. 15. The histogram represents data from a normal distribution because it’s bellshaped. 16. The histogram does not represent data from a normal distribution because it’s skewed right. 17. (Area left of z 1.2兲 共Area left of z 0兲 0.8849 0.5 0.3849 18. 共Area left of z 0兲 共Area left of z 2.25兲 0.5 0.0122 0.4878 19. (Area left of z 1.5兲 共Area left of z 0.5兲 0.9332 0.3085 0.6247 20. 共Area right of z 2兲 1 共Area left of z 2兲 1 0.9772 0.0228 21. 0.9131
22. 0.5319
23. 0.975
24. 0.8997
25. 1 0.2578 0.7422
26. 1 0.0256 0.9744
27. 1 0.8997 0.1003
28. 1 0.9994 0.0006
29. 0.005
30. 0.0008
31. 1 0.9469 0.0531
32. 1 0.9940 0.006
33. 0.9382 0.5 0.4382
34. 0.9979 0.5 0.4979
35. 0.5 0.0630 0.437
36. 0.5 0.3050 0.195
37. 0.9750 0.0250 0.95
38. 0.9901 0.0099 0.9802
39. 0.1003 0.1003 0.2006
40. 0.0250 0.0250 0.05
Light Bulb Life Spans
41. (a)
It is reasonable to assume that the life span is normally distributed because the histogram is nearly symmetric and bellshaped.
Frequency
f 7 6 5 4 3 2 1 x 1279 1626 1973 2320 2667
Hours
(b) x 1941.35 s ⬇ 432.385 (c) The sample mean of 1941.35 hours is less than the claimed mean, so on the average the bulbs in the sample lasted for a shorter time. The sample standard deviation of 432 hours is greater than the claimed standard deviation, so the bulbs in the sample had a greater variation in life span than the manufacturer’s claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 5
42. (a)

NORMAL PROBABILITY DISTRIBUTIONS
Heights of Males 8 7 6 5 4 3 2 1
x
63.85 65.85 67.85 69.85 71.85 73.85 75.85
Frequency
f
Inches
It is reasonable to assume that the heights are normally distributed because the histogram is nearly symmetric and bellshaped. (b) x ⬇ 68.75, s ⬇ 2.85 (c) The mean of your sample is 0.85 inch less than that of the previous study so the average height from the sample is less than in the previous study. The standard deviation is about 0.15 less than that of the previous study, so the heights are slightly less spread out than in the previous study. 43. (a) A 92.994
B 93.004
C 93.014
D 93.018
(b) x 93.014 ⇒ z
x 93.014 93.01 0.8 0.005
x 93.018 ⇒ z
x 93.018 93.01 1.6 0.005
x 93.004 ⇒ z
x 93.004 93.01 1.2 0.005
x 92.994 ⇒ z
x 92.994 93.01 3.2 0.005
(c) x 92.994 is unusual due to a relatively small zscore 共3.2兲. 44. (a) A 328
B 330
(b) x 328 ⇒ z
C 338
D 341
x 328 336 2.29 3.5
x 338 ⇒ z
x 338 336 0.57 3.5
x 330 ⇒ z
x 330 336 1.71 3.5
x 341 ⇒ z
x 341 336 1.43 3.5
(c) x 328 is unusual due to a relatively small zscore 共2.29兲.
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121
122
CHAPTER 5

45. (a) A 1186
NORMAL PROBABILITY DISTRIBUTIONS
B 1406
(b) x 1406 ⇒ z
C 1848
D 2177
x 1406 1518 0.36 308
x 1848 ⇒ z
x 1848 1518 1.07 308
x 2177 ⇒ z
x 2177 1518 2.14 308
x 1186 ⇒ z
x 1186 1518 1.08 308
(c) x 2177 is unusual due to a relatively large zscore 共2.14兲. 46. (a) A 14
B 18
(b) x 18 ⇒ z
C 25
D 32
x 18 21 0.63 4.8
x 32 ⇒ z
x 32 21 2.29 4.8
x 14 ⇒ z
x 14 21 1.46 4.8
x 25 ⇒ z
x 25 21 0.83 4.8
(c) x 32 is unusual due to a relatively large zscore 共2.29兲. 47. 0.6915
48. 0.1587
49. 1 0.95 0.05
50. 1 0.1003 0.8997
51. 0.8413 0.3085 0.5328
52. 0.9772 0.6915 0.2857
53. P共z < 1.45兲 0.9265
54. P共z < 0.45兲 0.6736
55. P共z > 0.95兲 1 P共z < 0.95兲 1 0.1711 0.8289 56. P共z > 1.85兲 1 P共z < 1.85兲 1 0.0322 0.9678 57. P共0.89 < z < 0兲 0.5 0.1867 0.3133 58. P共2.08 < z < 0兲 0.5 0.0188 0.4812 59. P共1.65 < z < 1.65兲 0.9505 0.0495 0.901 60. P共1.54 < z < 1.54兲 0.9382 0.0618 0.8764 61. P共z < 2.58 or z > 2.58兲 2共0.0049兲 0.0098 62. P共z < 1.54 or z > 1.54兲 2共0.618兲 0.1236
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
63.
36
48
60
72
84
The normal distribution curve is centered at its mean (60) and has 2 points of inflection (48 and 72) representing ± . 64.
350 400 450 500 550
The normal distribution curve is centered at its mean (450) and has 2 points of inflection (400 and 500) representing ± . 65. (a) Area under curve area of rectangle (base)(height) 共1兲共1兲 1 (b) P共0.25 < x < 0.5兲 共base兲共height兲 共0.25兲共1兲 0.25 (c) P共0.3 < x < 0.7兲 共base兲共height兲 共0.4兲共1兲 0.4 66. (a)
Area under curve Area of rectangle
f(x)
共base兲共height兲
0.10
共20 10兲 共0.10兲 0.05
1 x 10
15
20
(b) P共12 < x < 15兲 共base兲共height兲 共3兲共0.1兲 0.3 (c) P共13 < x < 18兲 共base兲共height兲 共5兲共0.1兲 0.5
5.2 NORMAL DISTRIBUTIONS: FINDIING PROBABILITIES
5.2 Try It Yourself Solutions 1a.
μ = 24
x = 28
x 20.8
24.0
27.2
Miles per gallon
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123
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CHAPTER 5
b. z

NORMAL PROBABILITY DISTRIBUTIONS
x 28 24 2.50 1.6
c. P共z < 2.50兲 0.9938 P共z > 2.50兲 1 0.9938 0.0062 d. The probability that a randomly selected manual transmission Focus will get more than 28 mpg in city driving is 0.0062. 2a. µ = 45
x 9
21
45
57
69 81
x = 33 x = 60 Time (in minutes)
b. z z
x 33 45 1 12 x 60 45 1.25 12
c. P共z < 1兲 0.1587 P共z > 1.25兲 0.8944 0.8944 0.1587 0.7357 d. If 150 shoppers enter the store, then you would expect 150共0.7357兲 ⬇ 110 shoppers to be in the store between 33 and 60 minutes. 3a. Read user’s guide for the technology tool. b. Enter the data. c. P共190 < x < 225兲 P共1 < z < 0.4兲 0.4968 The probability that a randomly selected U.S. man’s cholesterol is between 190 and 225 is about 0.4968.
5.2 EXERCISE SOLUTIONS 1. P共x < 80兲 P共z < 1.2兲 0.1151 2. P共x < 100兲 P共z < 2.8兲 0.9974 3. P共x > 92兲 P共z > 1.2兲 1 0.8849 0.1151 4. P共x > 75兲 P共z > 2.2兲 1 0.0139 0.9861 5. P共70 < x < 80兲 P共3.2 < z < 1.2兲 0.1151 0.0007 0.1144 6. P共85 < x < 95兲 P共0.2 < z < 1.8兲 0.9641 0.4207 0.5434 7. P共200 < x < 450兲 P共2.68 < z < 0.47兲 0.3192 0.0037 0.3155 8. P共670 < x < 800兲 P共1.32 < z < 2.45兲 0.9929 0.9066 0.0863 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
9. P共200 < x < 239兲 P共0.39 < z < 1.48兲 0.9306 0.6517 0.2789 10. P共200 < x < 239兲 P共0.46 < z < 0.48兲 0.6844 0.3228 0.3616 11. P共141 < x < 151兲 P共0.84 < z < 2.94兲 0.9984 0.7995 0.1989 12. P共140 < x < 149兲 P共1.74 < z < 0兲 0.5 0.0409 0.4591 13. (a) P共x < 66兲 P共z < 1.2兲 0.1151 (b) P共66 < x < 72兲 P共1.2 < z < 0.8兲 0.7881 0.1151 0.6730 (c) P共x > 72兲 P共z > 0.8兲 1 P共z < 0.8兲 1 0.7881 0.2119 14. (a) P共x < 7兲 P共z < 1.5兲 0.0668 (b) P共7 < x < 15兲 P共1.5 < z < 2.5兲 0.9938 0.0668 0.9270 (c) P共x > 15兲 P共z > 2.5兲 1 P共z < 2.5兲 1 0.9938 0.0062 15. (a) P共x < 17兲 P共z < 1.67兲 0.0475 (b) P共20 < x < 29兲 P共0.98 < z < 1.12兲 0.8686 0.1635 0.7051 (c) P共x > 32兲 P共z > 1.81兲 1 P共z < 1.81兲 1 0.9649 0.0351 16. (a) P共x < 23兲 P共z < 0.67兲 0.2514 (b) P共23 < x < 25兲 P共0.67 < z < 0兲 0.5 0.2514 0.2486 (c) P共x > 27兲 P共z > 0.67兲 1 P共z < 0.67兲 1 0.7486 0.2514 17. (a) P共x < 5兲 P共z < 2兲 0.0228 (b) P共5.5 < x < 9.5兲 P共1.5 < z < 2.5兲 0.9938 0.0668 0.927 (c) P共x > 10兲 P共z > 3兲 1 P共z < 3兲 1 0.9987 0.0013 18. (a) P共x < 70兲 P共z < 2.5兲 0.0062 (b) P共90 < x < 120兲 P共0.83 < z < 1.67兲 0.9525 0.2033 0.7492 (c) P共x > 140兲 P共z > 3.33兲 1 P共z < 3.33兲 1 0.9996 0.0004 19. (a) P共x < 4兲 共z < 2.44兲 0.0073 (b) P共5 < x < 7兲 P共1.33 < z < 0.89兲 0.8133 0.0918 0.7215 (c) P共x > 8兲 P共z > 2兲 1 0.9772 0.0228 20. (a) P共x < 17兲 P共z < 0.6兲 0.2743 (b) P共20 < x < 28兲 P共0 < z < 1.6兲 0.9452 0.5 0.4452 (c) P共x > 30兲 P共z > 2兲 1 0.9772 0.0228 21. (a) P共x < 600兲 P共z < 0.86兲 0.8051 ⇒ 80.51% (b) P共x > 550兲 P共z > 0.42兲 1 P共z < 0.42兲 1 0.6628 0.3372
共1000兲共0.3372兲 337.2 ⇒ 337 scores 22. (a) P共x < 500兲 P共z < 0.16兲 0.4364 ⇒ 43.64% (b) P共x > 600兲 P共z > 0.71兲 1 P共z < 0.71兲 1 0.7611 0.2389 共1500兲共0.2389兲 358.35 ⇒ 358 scores © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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23. (a) P共x < 200兲 P共z < 0.39兲 0.6517 ⇒ 65.17% (b) P共x > 240兲 P共z > 1.51兲 1 P共z < 1.51兲 1 0.9345 0.0655
共250兲共0.0655兲 16.375 ⇒ 16 women 24. (a) P共x < 239兲 P共z < 0.48兲 0.6844 (b) P共x > 200兲 P共z > 0.46兲 1 P共z < 0.46兲 1 0.3228 0.6772 共200兲共0.6672兲 135.44 ⇒ 135 women 25. (a) P共x > 11兲 P共z > 0.5兲 1 P共z < 0.5兲 1 0.6915 0.3085 ⇒ 30.85% (b) P共x < 8兲 P共z < 1兲 0.1587
共200兲共0.1587兲 31.74 ⇒ 32 fish 26. (a) P共x > 30兲 P共z > 1.67兲 1 P共z < 1.67兲 1 0.9525 0.0475 ⇒ 4.75% (b) P共x < 22兲 P共z < 1兲 0.1587 共50兲共0.1587兲 7.935 ⇒ 8 beagles 27. (a) P共x > 4兲 P共z > 3兲 1 P共z < 3兲 1 0.0013 0.9987 ⇒ 99.87% (b) P共x < 5兲 P共z < 2兲 0.0228
共35兲共0.0228兲 0.798 ⇒ 1 adult 28. (a) P共x > 125兲 P共z > 2.08兲 1 P共z < 2.08兲 1 0.9812 0.0188 → 1.88% (b) P共x < 90兲 P共z < 0.83兲 0.2033 共300兲共0.2033兲 60.99 ⇒ 61 bills 29. P共x > 2065兲 P共z > 2.17兲 1 P共z < 2.17兲 1 0.9850 0.0150 ⇒ 1.5% It is unusual for a battery to have a life span that is more than 2065 hours because of the relatively large zscore (2.17). 30. P共x < 3.1兲 P共z < 1.56兲 0.0594 ⇒ 5.94% It is not unusual for a person to consume less than 3.1 pounds of peanuts because the zscore is within 2 standard deviations of the mean. 31. Out of control, because the 10th observation plotted beyond 3 standard deviations. 32. Out of control, because two out of three consecutive points lie more than 2 standard deviations from the mean. (8th and 10th observations.) 33. Out of control, because the first nine observations lie below the mean and since two out of three consecutive points lie more than 2 standard deviations from the mean. 34. In control, because none of the three warning signals detected a change.
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
5.3 NORMAL DISTRIBUTIONS: FINDING VALUES 5.3 Try It Yourself Solutions 1ab. (1)
(2) 1 − 0.9616 = 0.0384
1 − 0.95 = 0.025 2
0.9616
z z
−z
0
c. (1) z 1.77
0.95
z 0
z
(2) z ± 1.96
2a. (1)
(2)
0.10
(3)
0.20
0.99
z
z
P10 0
z
P20 0
0
P99
b. (1) use area 0.1003
(2) use area 0.2005
(3) use area 0.9901
c. (1) z 1.28
(2) z 0.84
(3) z 2.33
3a. 70, 8 b. z 0.75 ⇒ x z 70 共0.75兲共8兲 64 z 4.29 ⇒ x z 70 共4.29兲共8兲 104.32 z 1.82 ⇒ x z 70 共1.82兲共8兲 55.44 c. 64 and 55.44 are below the mean. 104.32 is above the mean. 4a.
1% −2.33
z 0
b. z 2.33 c. x z 142 共2.33兲共6.51兲 ⬇ 126.83 d. So, the longest braking distance a Honda Accord could have and still be in the top 1% is 127 feet. 5a.
10%
−1.28
z 0
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b. z 1.28 c. x z 11.2 共1.28兲共2.1兲 8.512 d. So, the maximum length of time an employee could have worked and still be laid off is 8 years.
5.3 EXERCISE SOLUTIONS 1. z 0.70
2. z 0.81
3. z 0.34
4. z 1.33
5. z 0.16
6. z 2.41
7. z 2.39
8. z 0.84
9. z 1.645
10. z 1.04
11. z 1.555
12. z 2.33
13. z 2.33
14. z 1.04
15. z 0.84
16. z 0.13
17. z 1.175
18. z 0.44
19. z 0.67
20. z 0
21. z 0.67
22. z 1.28
23. z 0.39
24. z 0.39
25. z 0.38
26. z 0.25
27. z 0.58
28. z 1.99
29. z ± 1.645
30. z ± 1.96
⇒ z 1.18
31.
32.
0.785
⇒ z 0.79
0.785
⇒ z 0.79
0.0119
z
z
0 z
0
⇒ z 1.18
33.
34.
0.9881
0.215 0.0119
z
z 0
z
0
⇒ z ± 1.28
35. 0.8000
⇒ z ± 2.575
36. 0.99
0.10
0.10 0.005
−z
z 0
z
−z
0.005 z 0
z
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CHAPTER 5
⇒ z ± 0.06
37. 0.05
38.
NORMAL PROBABILITY DISTRIBUTIONS
0.12
0.44 0.475

⇒ z ± 0.15
0.44
0.475
−z 0 z
z
−z 0 z
39. (a) 95th percentile ⇒ Area 0.95 ⇒ z 1.645 x z 64.1 共1.645兲共2.71兲 68.56 inches (b) 1st quartile ⇒ Area 0.25 ⇒ z 0.67 x z 64.1 共0.67兲共2.71兲 62.28 inches 40. (a) 90th percentile ⇒ Area 0.90 ⇒ z 1.28 x z 69.6 共1.28兲共3.0兲 ⬇ 73.44 inches (b) 1st quartile ⇒ Area 0.25 ⇒ z 0.67 x z 69.6 共0.67兲共3.0兲 ⬇ 67.59 inches 41. (a) 10th percentile ⇒ Area 0.10 ⇒ z 1.28 x z 17.1 共1.28兲共4兲 11.98 pounds (b) 3rd quartile ⇒ Area 0.75 ⇒ z 0.67 x z 17.1 共0.67兲共4兲 19.78 pounds 42. (a) 5th percentile ⇒ Area 0.05 ⇒ z 1.645 x z 11.4 共1.645兲共3兲 6.47 pounds (b) 3rd quartile ⇒ Area 0.75 ⇒ z 0.67 x z 11.4 共0.67兲共3兲 13.41 pounds 43. (a) Top 30% ⇒ Area 0.70 ⇒ z 0.52 x z 127 共0.52兲共23.5) 139.22 days (b) Bottom 10% ⇒ Area 0.10 ⇒ z 1.28 x z 127 共1.28兲共23.5兲 96.92 days 44. (a) Top 25% ⇒ Area 0.75 ⇒ z 0.67 x z 15.4 共0.67兲共2.5兲 17.08 pounds (b) Bottom 15% ⇒ Area 0.15 ⇒ z 1.04 x z 15.4 共1.04兲共2.5兲 12.80 pounds 45. Lower 5% ⇒ Area 0.05 ⇒ z 1.645 x z 20 共1.645兲共0.07兲 19.88 46. Upper 7.5% ⇒ Area 0.955 ⇒ z 1.70 x z 32 共1.70兲共0.36兲 32.61
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47. Bottom 10% ⇒ Area 0.10 ⇒ z 1.28 x z 30,000 共1.28兲共2500兲 26,800 Tires which wear out by 26,800 miles will be replaced free of charge. 48. A: Top 10% ⇒ Area 0.90 ⇒ z 1.28 x z 72 共1.28兲共9兲 83.52 B: Top 30% ⇒ Area 0.70 ⇒ z 0.52 x z 72 共0.52兲共9兲 76.68 C: Top 70% ⇒ Area 0.30 ⇒ z 0.52 x z 72 共0.52兲共9兲 67.32 D: Top 90% ⇒ Area 0.10 ⇒ z 1.28 x z 72 共1.28兲共9兲 60.48 49. Top 1% ⇒ Area 0.99 ⇒ z 2.33 x z ⇒ 8 共2.33兲共0.03兲 ⇒ 7.930 ounces
5.4 SAMPLING DISTRIBUTIONS AND THE CENTRAL LIMIT THEOREM
5.4 Try It Yourself Solutions 1a.
b.
Sample
Mean
Sample
Mean
Sample
Mean
Sample
Mean
1, 1, 1 1, 1, 3 1, 1, 5 1, 1, 7 1, 3, 1 1, 3, 3 1, 3, 5 1, 3, 7 1, 5, 1 1, 5, 3 1, 5, 5 1, 5, 7 1, 7, 1 1, 7, 3 1, 7, 5 1, 7, 7
1 1.67 2.33 3 1.67 2.33 3 3.67 2.33 3 3.67 4.33 3 3.67 4.33 5
3, 1, 1 3, 1, 3 3, 1, 5 3, 1, 7 3, 3, 1 3, 3, 3 3, 3, 5 3, 3, 7 3, 5, 1 3, 5, 3 3, 5, 5 3, 5, 7 3, 7, 1 3, 7, 3 3, 7, 5 3, 7, 7
1.67 2.33 3 3.67 2.33 3 3.67 4.33 3 3.67 4.33 5 3.67 4.33 5 5.67
5, 1, 1 5, 1, 3 5, 1, 5 5, 1, 7 5, 3, 1 5, 3, 3 5, 3, 5 5, 3, 7 5, 5, 1 5, 5, 3 5, 5, 5 5, 5, 7 5, 7, 1 5, 7, 3 5, 7, 5 5, 7, 7
2.33 3 3.67 4.33 3 3.67 4.33 5 3.67 4.33 5 5.67 4.33 5 5.67 6.33
7, 1, 1 7, 1, 3 7, 1, 5 7, 1, 7 7, 3, 1 7, 3, 3 7, 3, 5 7, 3, 7 7, 5, 1 7, 5, 3 7, 5, 5 7, 5, 7 7, 7, 1 7, 7, 3 7, 7, 5 7, 7, 7
3 3.67 4.33 5 3.67 4.33 5 5.67 4.33 5 5.67 6.33 5 5.67 6.33 7
x
f
Probability
1 1.67 2.33 3 3.67 4.33 5 5.67 6.33 7
1 3 6 10 12 12 10 6 3 1
0.0156 0.0469 0.0938 0.1563 0.1875 0.1875 0.1563 0.0938 0.0469 0.0156
c. x 4,
2x
2 5 1.667, n 3
x
冪5 1.291 冪n 冪3
x 4, 2x ⬇ 1.667, x ⬇ 1.291 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 5
2a. x 64, x
冪n
9 冪100

NORMAL PROBABILITY DISTRIBUTIONS
0.9
b. n 100 n = 100
n = 36
x 59.5 61.0 62.5 64.0 65.5 67.0 68.5
Mean of phone bills (in dollars)
c. With a larger sample size, the mean stays the same but the standard deviation decreases. 3a. x 3.5, x
0.2 0.05 冪n 冪16
b.
x 3.35 3.40 3.45 3.50 3.55 3.60 3.65
Mean diameter (in feet)
4a. x 25, x
1.5 0.15 冪n 冪100
x 24.70
25.00
25.30
Mean time (in minutes)
b. x 24.7: z
x 25.5: z
x 24.7 25 0.3 2 1.5 0.15 冪n 冪100 x 25.5 25 0.5 3.33 1.5 0.15 冪n 冪100
c. P共z < 2兲 ⬇ 0.0228 P共z < 3.33兲 0.9996 P共8.7 < x < 9.5兲 P共2 < z < 3.33兲 0.9996 0.0228 0.9768
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132

CHAPTER 5
NORMAL PROBABILITY DISTRIBUTIONS
5a. x 306,258, x
冪n
44,000 ⬇ 12,701.7 冪12
x 280,855
306,258
331,661
Mean sales price (in dollars)
b. x 280,000: z
x 280,000 306,258 26,258 2.07 44,000 12,701.7 冪n 冪12
c. P共x > 200,000兲 P共z > 2.07兲 1 P共z < 2.07兲 1 0.0192 0.9808 6a. x 700: z x 700: z
x 700 625 0.5 150 x 700 625 75 1.58 150 47.43 冪n 冪10
b. P共z < 0.5兲 0.6915 P共z < 1.58兲 0.9429 c. There is a 69% chance an individual receiver will cost less than $700. There is a 94% chance that the mean of a sample of 10 receivers is less than $700.
5.4 EXERCISE SOLUTIONS 1. x 100
x
冪n
2. x 100
15 冪50
2.121
3. x 100
x
冪n
x
冪n
15 冪100
1.5
4. x 100 15
冪250
0.949
x
冪n
15 冪1000
0.474
5. False. As the size of the sample increases, the mean of the distribution of the sample mean does not change. 6. False. As the size of a sample increases the standard deviation of the distribution of sample means decreases. 7. False. The shape of the sampling distribution of sample means is normal for large sample sizes even if the shape of the population is nonnormal. 8. True
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
9. x 3.5, x 1.708
3.5, 2.958 The means are equal but the standard deviation of the sampling distribution is smaller. Sample
Mean
Sample
Mean
Sample
Mean
Sample
Mean
0, 0, 0 0, 0, 2 0, 0, 4 0, 0, 8 0, 2, 0 0, 2, 2 0, 2, 4 0, 2, 8 0, 4, 0 0, 4, 2 0, 4, 4 0, 4, 8 0, 8, 0 0, 8, 2 0, 8, 4 0, 8, 8
0 0.67 1.33 2.67 0.67 1.33 2 3.33 1.33 2 2.67 4 2.67 3.33 4 5.33
2, 0, 0 2, 0, 2 2, 0, 4 2, 0, 8 2, 2, 0 2, 2, 2 2, 2, 4 2, 2, 8 2, 4, 0 2, 4, 2 2, 4, 4 2, 4, 8 2, 8, 0 2, 8, 2 2, 8, 4 2, 8, 8
0.67 1.33 2 3.33 1.33 2 2.67 4 2 2.67 3.33 4.67 3.33 4 4.67 6
4, 0, 0 4, 0, 2 4, 0, 4 4, 0, 8 4, 2, 0 4, 2, 2 4, 2, 4 4, 2, 8 4, 4, 0 4, 4, 2 4, 4, 4 4, 4, 8 4, 8, 0 4, 8, 2 4, 8, 4 4, 8, 8
1.33 2 2.67 4 2 2.67 3.33 4.67 2.67 3.33 4 5.33 4 4.67 5.33 6.67
8, 0, 0 8, 0, 2 8, 0, 4 8, 0, 8 8, 2, 0 8, 2, 2 8, 2, 4 8, 2, 8 8, 4, 0 8, 4, 2 8, 4, 4 8, 4, 8 8, 8, 0 8, 8, 2 8, 8, 4 8, 8, 8
2.67 3.33 4 5.33 3.33 4 4.67 6 4 4.67 5.33 6.67 5.33 6 6.67 8
10. 再120 120, 120 140, 120 180, 120 220, 140 120, 140 140, 140 180, 140 220, 180 120, 180 140, 180 180, 180 220, 220 120, 220 140, 220 180, 220 220冎
x 165, x 27.157 165, 38.406 The means are equal but the standard deviation of the sampling distribution is smaller. 11. (c) Because x 16.5, x 12. (b) Because x 5.8, x 13. z
冪n
11.9 冪100
1.19 and the graph approximates a normal curve.
2.3 ⬇ 0.23 and the graph approximates a normal curve. 冪n 冪100
x 12.2 12 0.2 1.26 0.95 0.158 冪n 冪36
P共x < 12.2兲 P共z < 1.26兲 0.8962 14. z
x 12.2 12 0.2 2.11 0.95 0.095 冪n 冪100
P共x > 12.2兲 P共z > 2.11兲 1 P共z < 2.11兲 1 0.9826 0.0174 The probability is unusual because it is less than 0.05. 15. z
x 221 220 1 2.22 39 0.450 冪n 冪75
P共x > 221兲 P共z > 2.22兲 1 P共z < 2.22兲 1 0.9868 0.0132 The probability is unusual because it is less than 0.05. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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16. z

NORMAL PROBABILITY DISTRIBUTIONS
x 12,753 12,750 3 10.59 1.7 0.283 冪n 冪36
P共x < 12,750 or x > 12,753兲 P共z < 0 or z > 10.59兲 0.5 0.000 0.5 17. x 87.5
x
18. x 800
6.25 ⬇ 1.804 冪n 冪12
100 ⬇ 25.820 冪n 冪15
x
x
x
82.1 83.9 85.7 87.5 89.3 91.1 92.9
722 748 774 800 826 852 878
Mean height (in feet)
Mean number of eggs
19. x 224
x
冪n
20. x 47.2
8
x
1.265
冪40
冪n
3.6 冪36
0.6
x 221.5
224
x
226.5
46
Mean price (in dollars)
48.4
Mean age (in years)
21. x 110
x
47.2
22. x 51.5
38.5 ⬇ 8.609 冪n 冪20
x
17.1 ⬇ 3.42 冪n 冪25
x 92.8
110
x
127.2
44.7
Mean consumption of red meat (in pounds)
23. x 87.5, x
x 87.5, x
51.5
58.3
Mean consumption (in gallons)
6.25 ⬇ 1.276 冪n 冪24 冪n
6.25 冪36
⬇ 1.042
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
n = 36 n = 24
n = 12 x 83.6 84.9 86.2 87.5 88.8 90.1 91.4
Mean height (in feet)
As the sample size increases, the standard error decreases, while the mean of the sample means remains constant. 24. x 800, x
x 800, x
100 ⬇ 18.257 冪n 冪30 冪n
100 冪45
⬇ 14.907
n = 45 n = 30
n = 15
As the sample size increases, the standard error decreases. 25. z
x 740 760 780 800 820 840 860
Mean number of eggs
x 44,000 46,700 2700 ⬇ 3.12 5600 864.10 冪n 冪42
P共x < 44,000兲 P共z < 3.12兲 0.0009 26. z
x 55,000 59,100 4100 ⬇ 14.27 1700 207.35 冪n 冪35
P共x < 55,000兲 P共z < 14.27兲 ⬇ 0 27. z
z
x 2.768 2.818 0.05 ⬇ 6.29 0.045 0.00795 冪n 冪32 x 2.918 2.818 0.1 ⬇ 12.57 0.045 0.00795 冪n 冪32
P共2.768 < x < 2.918兲 P共6.29 < z < 12.57兲 ⬇ 1 0 1 28. z
z
x 3.310 3.305 0.005 ⬇ 0.63 0.049 0.00794 冪n 冪38 x 3.320 3.305 0.015 ⬇ 1.89 0.049 0.00794 冪38 冪n
P共3.310 < x < 3.320兲 P共0.63 < z < 1.89兲 0.9706 0.7357 0.2349 29. z
x 66 64.1 1.9 ⬇ 5.43 2.71 0.350 冪n 冪60
P共x > 66兲 P共z > 5.43兲 ⬇ 0 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
135
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CHAPTER 5
30. z

NORMAL PROBABILITY DISTRIBUTIONS
x 70 69.6 0.4 ⬇ 1.03 3.0 0.387 冪n 冪60
P共x > 70兲 P共z > 1.03兲 1 P共z < 1.03兲 1 0.8485 0.1515 x 70 64.1 ⬇ 2.18 2.71 P共x < 70兲 P共z < 2.18兲 0.9854
31. z
z
x 70 64.1 5.9 ⬇ 9.73 2.71 0.606 冪n 冪20
P共x < 70兲 P共z < 9.73兲 ⬇ 1 It is more likely to select a sample of 20 women with a mean height less than 70 inches, because the sample of 20 has a higher probability. 32. z
x 65 69.6 ⬇ 1.53 3.0
P共x < 65兲 P共z < 1.53兲 0.0630 z
x 65 69.6 4.6 ⬇ 5.94 3.0 0.775 冪n 冪15
P共x < 65兲 P共z < 5.94兲 ⬇ 0 It is more likely to select one man with a height less than 65 inches because the probability is greater. 33. z
x 127.9 128 0.1 ⬇ 3.16 0.20 0.032 冪n 冪40
P共x < 127.9兲 P共z < 3.16兲 0.0008 Yes, it is very unlikely that we would have randomly sampled 40 cans with a mean equal to 127.9 ounces, because it is more than 2 standard deviations from the mean of the sample means. 34. z
x 64.05 64 0.05 ⬇ 2.87 0.11 0.017 冪n 冪40
P共x > 64.05兲 P共z > 2.87兲 1 P共z < 2.87兲 1 0.9979 0.0021 Yes, it is very unlikely that we would have randomly sampled 40 containers with a mean equal to 64.05 ounces, because it is more then 2 standard deviations from the mean of the sample means. 35. (a) 96
0.5 z
x 96.25 96 0.25 ⬇ 3.16 0.5 0.079 冪 冪90
P共x 96.25兲 P共z > 3.16兲 1 P共z < 3.16兲 1 0.9992 0.0008 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
(b) Claim is inaccurate. (c) Assuming the distribution is normally distributed: z
x 96.25 96 0.5 0.5
P共x > 96.25兲 P共z > 0.5兲 1 P共z < 0.5兲 1 0.6915 0.3085 Assuming the manufacturer’s claim is true, an individual board with a length of 96.25 would not be unusual. It is within 1 standard deviation of the mean for an individual board. 36. (a) 10
0.5 z
x 10.21 10 0.21 2.1 0.5 0.1 冪n 冪25
P共x 10.21兲 P共z 2.1兲 1 P共z 2.1兲 1 0.9821 0.0179 (b) Claim is inaccurate. (c) Assuming the distribution is normally distributed: z
x 10.21 10 0.42 0.5
P共x 10.21兲 P共z 0.42兲 1 P共z 0.42兲 1 0.6628 0.3372 Assuming the manufacturer’s claim is true, an individual carton with a weight of 10.21 would not be unusual because it is within 1 standard deviation of the mean for an individual ice cream carton. 37. (a) 50,000
800 z
x 49,721 50,000 279 3.49 800 80 冪n 冪100
P共x 49,721兲 P共z 3.49兲 0.0002 (b) The manufacturer’s claim is inaccurate. (c) Assuming the distribution is normally distributed: z
x 49,721 50,000 0.35 800
P共x < 49,721兲 P共z < 0.35兲 0.3669 Assuming the manufacturer’s claim is true, an individual tire with a life span of 49,721 miles is not unusual. It is within 1 standard deviation of the mean for an individual tire.
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38. (a) 38,000
1000 x 37,650 38,000 350 2.47 1000 141.42 冪n 冪50
z
P共x 37,650兲 P共z 2.47兲 0.0068 (b) Claim is inaccurate. (c) Assuming the distribution is normally distributed: x 37,650 38,000 0.35 1000
z
P共x < 37,650兲 P共z < 0.35兲 0.3632 Assuming the manufacturer’s claim is true, an individual brake pad lasting less than 37,650 miles would not be unusual, because it is within 1 standard deviation of the mean for an individual brake pad. 39. 518
115 z
x 530 518 12 0.74 115 16.26 冪n 冪50
P共x 530兲 P共z 0.74兲 1 P共z 0.74兲 1 0.7704 0.2296 The high school’s claim is not justified because it is not rare to find a sample mean as large as 530. 40. 4
0.5 z
x 4.2 4 0.2 4 0.5 0.05 冪100 冪n
P共x 4.2兲 P共z 4兲 1 P共z 4兲 ⬇ 1 1 ⬇ 0 It is very unlikely the machine is calibrated to produce a bolt with a mean of 4 inches. 41. Use the finite correction factor since n 55 > 40 0.05N. x
2.871 2.876 0.005 ⬇ 4.27 0.009 共0.00121兲冪0.9324
n 800 55 冪55 N1 800 1 P共x < 2.871兲 P共z < 4.27兲 ⬇ 0
z
冪n
冪
冪
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
42. Use the finite correction factor since n 30 > 25 0.05N. z
z
冪n 冪n
x
冪 N n1 x
冪 N n1
2.5 3.32 0.82 ⬇ 4.25 1.09 共0.199兲冪0.9419 500 30 冪30 500 1
冪
4 3.32 1.09 冪30
30 冪500 500 1
0.68 ⬇ 3.52 共0.199兲冪0.9419
P共2.5 < x < 4兲 P共4.25 < z < 3.52兲 ⬇ 1 0 1
45.
Sample
Number of boys
Proportion
bbb
3
1
bbg
2
2 3
bgb
2
2 3
bgg
1
1 3
gbb
2
2 3
gbg
1
1 3
ggb
1
1 3
ggg
0
0
Sample
Sample Mean
bbb
1
bbg
2 3
bgb
2 3
bgg
1 3
gbb
2 3
gbg
1 3
ggb
1 3
ggg
0
44.
Proportion
Probability
0
1 8
1 3
3 8
2 3
3 8
1
1 8
Proportion of Boys from Three Births 0.500
Probability
43.
0.375 0.250 0.125 0
1 3
2 3
1
Sample proportions
The spread of the histogram for each proportion is equal to the number of occurences of 0, 1, 2, and 3 boys, respectively, from the binomial distribution.
The sample mean is the same as the proportion of boys in each sample.
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140
46.
CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
Sample
Number of boys
Proportion
bbbb
4
1
bbbg
3
bbgb
Sample
Number of boys
Proportion
Proportion
Probability
gbbb
3
3 4
0
1 16
3 4
gbbg
2
1 2
1 4
1 4
3
3 4
gbgb
2
1 2
1 2
3 8
bbgg
2
1 2
gbgg
1
1 4
3 4
1 4
bgbb
3
3 4
ggbb
2
1 2
1
1 16
bgbg
2
1 2
ggbg
1
1 4
bggb
2
1 2
gggb
1
1 4
bggg
1
1 4
gggg
0
0
Probability
Proportion of Boys from Four Births 0.3750 0.3125 0.2500 0.1875 0.1250 0.0625 0
1 4
1 2
3 4
1
Sample proportions
p p 0.70 0.75 0.05 1.10 0.0456 0.75共0.25兲 pq 90 n ^
47. z
冪
冪
P共p < 0.70兲 P共z < 1.10兲 0.1357
5.5 NORMAL APPROXIMATIONS T0 BINOMIAL DISTRIBUTIONS
5.5 Try It Yourself Solutions 1a. n 70, p 0.80, q 0.20 b. np 56, nq 14 c. Because np 5 and nq 5, the normal distribution can be used. d. np 共70兲共0.80兲 56
冪npq 冪共70兲共0.80兲共0.20兲 ⬇ 3.35 2a. (1) 57, 58, . . . , 83
(2) . . . , 52, 53, 54
b. (1) 56.5 < x < 83.5 (2) x < 54.5 3a. n 70, p 0.80 np 56 5 and nq 14 5 The normal distribution can be used.
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
b. np 56
冪npq ⬇ 3.35 c. x > 50.5 µ = 56
x = 50.5 x 47
50
53
56
59
62
65
Number responding yes
d. z
x 50.5 56 ⬇ 1.64 3.35
e. P共z < 1.64兲 0.0505 P共x > 50.5兲 P共z > 1.64兲 1 P共z < 1.64兲 0.9495 The probability that more than 50 respond yes is 0.9495. 4a. n 200, p 0.38 np 76 5 and nq 124 5 The normal distribution can be used. b. np 76
冪npq ⬇ 6.86 c. P共x 85.5兲 µ = 76
x = 85.5 x 55 62 69 76 83 90 97
Number responding yes
d. z
x 85.5 76 ⬇ 1.38 6.86
e. P共x < 65.5兲 P共z < 1.38兲 0.9162 The probability that at most 65 people will say yes is 0.9162. 5a. n 200, p 0.86 np 172 5 and nq 28 5 The normal distribution can be used. b. np 172
冪npq ⬇ 4.91 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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c. P(169.5 < x < 170.5兲 µ = 172 x = 169.5
x = 170.5
x 157 162 167 172 175 177 182
Number responding yes
d. z
x 169.5 172 ⬇ 0.51 4.91
z
x 170.5 172 ⬇ 0.31 4.91
e. P共z < 0.51兲 0.3050 P共z < 0.31兲 0.3783 P共0.51 < z < 0.31兲 0.3783 0.3050 0.0733 The probability that exactly 170 people will respond yes is 0.0733.
5.5 EXERCISE SOLUTIONS 1. np 共24兲共0.85兲 20.4 5
2. np 共15兲共0.70兲 10.5 5
nq 共24兲共0.15兲 3.6 < 5
nq 共15兲共0.30兲 4.5 < 5
Cannot use normal distribution.
Cannot use the normal distribution.
3. np (18兲共0.90兲 16.2 5
4. np 共20兲共0.65兲 13 5
nq 共18兲共0.10兲 1.8 < 5
nq 共20兲共0.35兲 7 5
Cannot use normal distribution.
Use the normal distribution.
5. n 10, p 0.85, q 0.15 np 8.5 > 5, nq 1.5 < 5 Cannot use normal distribution because nq < 5. 6. n 20, p 0.63, q 0.37 np 12.6 5, nq 7.4 5 Use the normal distribution.
np 共20兲共0.66兲 13.2 冪npq 冪共20兲共0.66兲共0.34兲 ⬇ 2.12 7. n 10, p 0.99, q 0.03
8. n 30, p 0.086, q 0.914
np 9.9 5, nq 0.1 < 5
np 2.58 < 5, nq 27.42 5
Cannot use normal distribution because nq < 5.
Cannot use the normal distribution because np < 5.
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CHAPTER 5
9. d
10. b
11. a
12. c
13. a
14. d
15. c
16. b

NORMAL PROBABILITY DISTRIBUTIONS
17. Binomial: P共5 x 7兲 0.162 0.198 0.189 0.549 Normal: P共4.5 x 7.5兲 P共0.97 < z < 0.56兲 0.7123 0.1660 0.5463 18. Binomial: P共2 x 4兲 P共2兲 P共3兲 P共4兲 ⬇ 0.016 0.054 0.121 0.191 Normal: P共1.5 x 4.5兲 P共2.60 z 0.87兲 0.1922 0.0047 0.1875 19. n 30, p 0.07 → np 2.1 and nq 27.9 Cannot use normal distribution because np < 5. (a) P共x 10兲 30C10共0.07兲10共0.93兲20 ⬇ 0.0000199 (b) P共x 10兲 1 P共x < 10兲 1 关30C0共0.07兲0共0.93兲30 30C1共0.07兲1共0.93兲29 . . . 30C9共0.07兲9共0.93兲21兴 1 .999977 ⬇ 0.000023 (c) P共x < 10兲 ⬇ 0.999977 (see part b) (d) n 100, p 0.07 → np 7 and nq 93 Use normal distribution. z
x 4.5 7 ⬇ 0.98 2.55
P共x < 5兲 P共x < 4.5兲 P共z < 0.98兲 0.1635
x = 4.5 x 1.8
7.0
12.2
Number of donors
20. n 32, p 0.34 → np 10.88 and nq 21.12 Use normal distribution. (a) z
x 11.5 10.88 ⬇ 0.23 2.68
z
x 12.5 10.88 ⬇ 0.60 2.68
P共x 12兲 P共11.5 < x < 12.5兲 P共0.23 < z < 0.60兲
x = 11.5
x = 12.5
x 2 4 6 8 10 12 14 16 18 20
Number of people
0.7257 0.5910 0.1347 (b) P共x 12兲 P共x > 11.5兲 P共z > 0.23兲 1 P共z < 0.23兲
x = 11.5
1 0.5910 0.4090
x 2 4 6 8 10 12 14 16 18 20
Number of people
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(c) P共x < 12兲 P共x < 11.5兲 P共z < 0.23兲 0.5910 x = 11.5
x 2 4 6 8 10 12 14 16 18 20
Number of people
(d) n 150, p 0.34 → np 51 and nq 99 z
x 59.5 51 ⬇ 1.47 5.80 x = 59.5
P共x < 60兲 P共x < 59.5兲 P共z < 1.47兲
x 32 36 40 44 48 52 56 60 64 68
Number of people
0.9292 21. n 250, p 0.05, q 0.95 np 12.5 5, nq 237.5 5 Use the normal distribution. (a) z
x 15.5 12.5 0.87 3.45
z
x 16.5 12.5 1.16 3.45
P共x 16兲 ⬇ P共15.5 x 16.5兲 P共0.87 z 1.16兲
x = 15.5
x = 16.5
x 5.6
12.5
19.4
Number of workers
0.8770 0.8078 0.0692 (b) P共x 9兲 ⬇ P共x 8.5兲 P共z 1.16) 1 P共z 1.16兲 1 0.1230 0.8770
x = 8.5
x 5.6
12.5
19.4
Number of workers
(c) P共x < 16兲 ⬇ P共x 15.5兲 P共z 0.87兲 0.8078
x = 15.5
x 5.6
12.5
19.4
Number of workers
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
(d) n 500, p 0.05, q 0.95 np 25 5, nq 475 5 Use normal distribution. z
x 29.5 25 0.92 4.87
P共x < 30兲 ⬇ P共x < 29.5兲 P共z < 0.92兲 0.8212
x = 29.5
x 15.3
25
34.7
Number of workers
22. n 50, p 0.32, q 0.68 np 16 5, nq 34 5 Use normal distribution. (a) z
x 11.5 16 1.36 3.30
z
x 12.5 16 1.06 3.30
P共x 12兲 ⬇ P共11.5 x 12.5兲 P共1.36 < z < 1.06兲
x = 11.5
x = 12.5
x 9.4
16
22.6
Number of workers
0.1446 0.0869 0.0577 (b) P共x 14兲 ⬇ P共x 13.5兲 P共z 0.76兲 1 P共z 0.76兲
x = 13.5
1 0.2236 0.7764
x 9.4
0.7764
16
22.6
Number of workers
(c) P共x < 18兲 ⬇ P共x 17.5兲 P共z 0.45兲 0.6736
x = 17.5
x 9.4
16
22.6
Number of workers
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(d) n 150, p 0.32, q 0.68 np 48, nq 102 x = 29.5
Use normal distribution. z
x 29.5 48 3.24 5.71
x 36.6
48
59.4
Number of workers
P共x < 30兲 ⬇ P共x 29.5兲 P共z 3.24兲 0.0006 23. n 40, p 0.52 → np 20.8 and nq 19.2 Use the normal distribution. (a) z
x 23.5 20.8 ⬇ 0.85 3.160
P共x 23兲 P共x < 23.5兲 P共z < 0.85兲 0.8023
x = 23.5
x 12 14 16 18 20 22 24 26 28 30
Number of people
(b) z
x 17.5 20.8 ⬇ 1.04 3.160
P共x 18兲 P共x > 17.5兲 P共z > 1.04兲 1 P共z < 1.04兲 1 0.1492 0.8508
x = 17.5
x 12 14 16 18 20 22 24 26 28 30
Number of people
(c) P共x > 20兲 P共x > 20.5兲 P共z > 0.09兲 1 P共z < 0.09兲 1 0.4641 0.5359
x = 20.5
x 12 14 16 18 20 22 24 26 28 30
Number of people
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
(d) n 650, p 0.52 → np 338 and nq 312 Use normal distribution. z
x 350.5 338 ⬇ 0.98 12.74
P共x > 350兲 P1(x > 350.5兲 P共z > 0.98兲 1 P共z < 0.98兲 1 0.8365 0.1635
x = 350.5 x 299 312 325 338 351 364 377
Number of people
24. n 10, p 0.029 → np 0.29 and nq 9.71 Cannot use normal distribution because np < 5. (a) P共x 3兲 P共x 0兲 P共x 1兲 P共x 2兲 P共x 3兲 10C0共0.029兲0共0.971兲10 10C1共0.029兲1共0.971兲9 10C2共0.029兲2共0.971兲8 10C3共0.029兲3共0.971兲7 ⬇ 0.9999 (b) P共x 1兲 1 P共x < 1兲 1 P共x 0兲 ⬇ 0.2549 (c) P共x > 2兲 1 P共x 2兲 ⬇ 1 0.9975 0.0025 (d) n 50, p 0.029 → np 1.45 and nq 48.55 Cannot use normal distribution. P共x 0兲 50C0共0.029兲0共0.971兲50 ⬇ 0.2296 25. (a) n 25, p 0.24, q 0.76 np 6 5, nq 19 5 Use normal distribution. (b) z
x 8.5 6 1.17 2.135
P共x > 8兲 ⬇ P共x 8.5兲 P共z 1.17兲 1 P共z 1.17兲 1 0.8790 0.121 (c) z
x 7.5 6 0.70 2.135
z
x 8.5 6 1.17 2.135
P共x 8兲 ⬇ P共7.5 x 8.5兲 P共0.07 z 1.17兲 0.8790 0.7580 0.121 It is not unusual for 8 out of 25 homeowners to say their home is too small because the zscore is within 1 standard deviation of the mean.
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26. (a) n 40, p 0.80, q 0.20 np 32 5, nq 8 5 Use normal distribution. (b) z
x 26.5 32 2.17 2.53
P共x 26兲 ⬇ P共x 26.5兲 P共z 2.17兲 0.0150 (c) z
x 25.5 32 2.57 2.53
z
x 26.5 32 2.17 2.53
P共x 26兲 ⬇ P共25.5 x 26.5兲 P共2.57 z 2.17兲 0.0150 0.0051 0.0099 Yes, because the zscore is more than two standard deviations from the mean. 27. n 250, p 0.70 60% say no → 250共0.6兲 150 say no while 100 say yes. z
x 99.5 175 10.41 7.25
P(less than 100 yes) P共x < 100兲 P共x < 99.5兲 P共z < 10.41兲 ⬇ 0 It is highly unlikely that 60% responded no. Answers will vary. 28. n 200, p 0.11 9% of 200 18 people z
x 17.5 22 ⬇ ⬇ 1.02 4.42
P共x < 18兲 P共x < 17.5兲 P共z < 1.02兲 0.1539 It is probable that 18 of the 200 people responded that they participate in hiking. Answers will vary. 29. n 100, p 0.75 z
x 69.5 75 ⬇ 1.27 4.33
P(reject claim) P共x < 70兲 P共x < 69.5兲 P共z < 1.27兲 0.1020 30. n 100, p 0.65 z
x 69.5 65 ⬇ 0.94 4.77
P共accept claim兲 P共x 70兲 P共x > 69.5兲 P共z > 0.94兲 1 P共z < 0.94兲 1 0.8264 0.1736
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
CHAPTER 5 REVIEW EXERCISE SOLUTIONS 1. 15, 3 3. x 1.32: z
2. 3, 5
x 1.32 1.5 2.25 0.08
x 1.54: z
x 1.54 1.5 0.5 0.08
x 1.66: z
x 1.66 1.5 2 0.08
x 1.78: z
x 1.78 1.5 3.5 0.08
4. 1.32 and 1.78 are unusual. 5. 0.6293
6. 0.9946
7. 0.3936
8. 0.9573
9. 1 0.9535 0.0465
10. 1 0.5478 0.4522
11. 0.5 0.0505 0.4495
12. 0.8508 0.0606 0.7902
13. 0.9564 0.5199 0.4365
14. 0.9750 0.0250 0.95
15. 0.0668 0.0668 0.1336
16. 0.7389 0.0154 0.7543
17. P共z < 1.28兲 0.8997
18. P共z > 0.74兲 0.7704
19. P共2.15 < x < 1.55兲 0.9394 0.0158 0.9236 20. P共0.42 < z < 3.15兲 0.9992 0.6628 0.3364 21. P共z < 2.50 or z > 2.50兲 2共0.0062兲 0.0124 22. P共z < 0 or z > 1.68兲 0.5 0.0465 0.5465 23. (a) z
x 1900 2200 ⬇ 0.48 625
P共x < 1900兲 P共z < 0.48兲 0.3156 (b) z
x 2000 2200 ⬇ 0.32 625
z
x 25,900 2200 ⬇ 0.48 625
P共2000 < x < 2500兲 P1(0.32 < z < 0.48兲 0.6844 0.3745 0.3099 x 2450 2200 0.4 625 P共x > 2450兲 P1(z > 0.4兲 0.3446
(c) z
24. (a) z z
x 1 1.5 2 0.25 x 2 1.5 2 0.25
P共1 < x < 2兲 P共2 < z < 2兲 0.9772 0.0228 0.9544
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149
150
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NORMAL PROBABILITY DISTRIBUTIONS
(b) z
x 1.6 1.5 0.4 0.25
z
x 2.2 1.5 2.8 0.25
P共1.6 < x < 2.2兲 P共0.4 < z < 2.8兲 0.9974 0.6554 0.3420 (c) z
x 2.2 1.5 2.8 0.25
P(x > 2.2兲 P共z > 2.8兲 0.0026 25. z 0.07
26. z 1.28
27. z 1.13
28. z 2.05
29. z 1.04
30. z 0.84
31. x z 52 共2.4兲共2.5兲 46 meters 32. x z 52 共1.2兲共2.5兲 55 meters 33. 95th percentile ⇒ Area 0.95 ⇒ z 1.645 x z 52 共1.645兲共2.5兲 56.1 meters 34. 3rd Quartile ⇒ Area 0.75 ⇒ z 0.67 x z 52 共0.67兲共2.5兲 53.7 35. Top 10% ⇒ Area 0.90 ⇒ z 1.28 x z 52 共1.28兲共2.5兲 55.2 meters 36. Bottom 5% ⇒ Area 0.05 ⇒ z 1.645 x z 52 共1.645兲共2.5兲 ⬇ 47.9 37. 再90 90 90, 90 90 120, 90 90 160, 90 90 210, 90 90 300, 90 120 90, 90 120 120, 90 120 160, 90 120 210, 90 120 300, 90 160 90, 90 160 120, 90 160 160, 90 160 210, 90 160 300, 90 210 90, 90 210 120, 90 210 160, 90 210 210, 90 210 300, 90 300 90, 90 300 120, 90 300 160, 90 300 210, 90 300 300, 120 90 90, 120 90 120, 120 90 160, 120 90 210, 120 90 300, 120 120 90, 120 120 120, 120 120 160, 120 120 210, 120 120 300, 120 160 90, 120 160 120, 120 160 160, 120 160 210, 120 160 300, 120 210 90, 120 210 120, 120 210 160, 120 210 210, 120 210 300, 120 300 90, 120 300 120, 120 300 160, 120 300 210, 120 300 300, 160 90 90, 160 90 120, 160 90 160, 160 90 210, 160 90 300, 160 120 90, 160 120 120, 160 120 160, 160 120 210, 160 120 300, 160 160 90, 160 160 120, 160 160 160, 160 160 210, 160 160 300, 160 210 90, 160 210 120, 160 210 160, 160 210 210, 160 210 300, 160 300 90, 160 300 120, 160 300 160, 160 300 210, 160 300 300, 210 90 90, 210 90 120, 210 90 160, 210 90 210, 210 90 300, 210 120 90, 210 120 120, 210 120 160, 210 120 210, 210 120 300, 210 160 90, 210 160 120, 210 160 160, 210 160 210, 210 160 300, 210 210 90, 210 210 120, 210 210 160, 210 210 210, 210 210 300, 210 300 90, 210 300 120, 210 300 160, 210 300 210, 210 300 300, 300 90 90, 300 90 120, 300 90 160, 300 90 210, 300 90 300, 300 120 90, 300 120 120, 300 120 160, 300 120 210, 300 120 300, 300 160 90, 300 160 120, 300 160 160, 300 160 210, 300 160 300, 300 210 90, 300 210 120, 300 210 160, 300 210 210, 300 210 300, 300 300 90, 300 300 120, 300 300 160, 300 300 210, 300 300 300冎
176, ⬇ 73.919 x 176, x ⬇ 42.677 The means are the same, but x is less than .
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
38. 再00, 01, 02, 03, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33冎
1.5, ⬇ 1.118 x 1.5, x ⬇ 0.791 The means are the same, but x is less than . 39. x 144.3, x
51.6 ⬇ 8.722 冪n 冪35
x 126.9
144.3
161.7
Mean consumption (in pounds)
40. x 218.2, x
68.1 ⬇ 10.767 冪n 冪40
x 196.7
218.2
239.7
Mean consumption (in pounds)
41. (a) z
x 1900 2200 300 ⬇ 1.66 625 180.42 冪n 冪12
P共x < 1900兲 P共z < 1.66兲 0.0485 (b) z
z
x 2000 2200 200 ⬇ 1.11 625 180.42 冪n 冪12 x 2500 2200 300 ⬇ 1.66 625 180.42 冪n 冪12
P共2000 < x < 2500兲 P共1.11 < z < 1.66兲 0.9515 0.1335 0.8180 (c) z
x 2450 2200 250 ⬇ 1.39 625 180.42 冪n 冪12
P共x > 2450兲 P共z > 1.39兲 0.0823 (a) and (c) are smaller, (b) is larger. This is to be expected because the standard error of the sample mean is smaller.
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151
152
CHAPTER 5
42. (a) z
z

NORMAL PROBABILITY DISTRIBUTIONS
x 1.0 1.5 0.5 ⬇ 5.29 0.25 0.0945 冪n 冪7 x 2.0 1.5 0.5 ⬇ 5.29 0.25 0.0945 冪n 冪7
P共1.0 < x < 2.0兲 P共5.29 < z < 5.29兲 ⬇ 1 (b) z
z
x 1.6 1.5 0.1 ⬇ 1.06 0.25 0.0945 冪n 冪7 x 2.2 1.5 1.7 ⬇ 7.41 0.25 0.0945 冪n 冪7
P共1.6 < x < 2.2兲 P共1.06 < z < 7.41兲 ⬇ 1 0.8554 0.1446 (c) z
x 2.2 1.5 1.7 ⬇ 7.41 0.25 0.0945 冪n 冪7
P共x > 2.2兲 P共z > 7.41兲 ⬇ 0 (a) is larger and (b) and (c) are smaller. 43. (a) z
x 29,000 29,200 200 ⬇ 0.89 1500 223.61 冪n 冪45
P共x < 29,000兲 P共z < 0.89兲 ⬇ 0.1867 (b) z
x 31,000 29,200 1500 ⬇ 8.05 1500 223.61 冪n 冪45
P共x > 31,000兲 P共z > 8.05兲 ⬇ 0 44. (a) z
x 1400 1300 100 ⬇ 2.4 250 41.67 冪n 冪36
P共x < 1400兲 P共z < 2.4兲 0.9918 (b) z
x 1150 1300 150 3.6 250 41.67 冪n 冪36
P共x > 1150兲 P共z > 3.6兲 ⬇ 1 0 0 45. Assuming the distribution is normally distributed: z
x 1.125 1.5 0.375 ⬇ 2.90 .5 0.129 冪n 冪15
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
46. Assuming the distribution is normally distributed: z
x 5250 5000 250 ⬇ 3.23 300 77.46 冪n 冪15
P共x > 5250兲 P共z > 3.23兲 1 P共z < 3.23兲 1 0.9994 .0006 47. n 12, p 0.95, q 0.05 np 11.4 > 5, but nq 0.6 < 5 Cannot use the normal distribution because nq < 5. 48. n 15, p 0.59, q 0.41 np 8.85 > 5 and nq 6.15 > 5 Use the normal distribution.
np 8.85, 冪npq 冪15共0.59兲共0.41兲 ⬇ 1.905 49. P共x 25兲 P共x > 24.5兲
50. P共x 36兲 P共x < 36.5兲
51. P共x 45兲 P共44.5 < x < 45.5兲
52. P共x 50兲 P共49.5 < x < 50.5兲
53. n 45, p 0.70 → np 31.5, nq 13.5 Use normal distribution.
np 31.5, 冪npq 冪45共0.70兲共0.30兲 ⬇ 3.07 z
x = 20.5
x 20.5 31.5 ⬇ 3.58 3.07
x 25.3
31.5
37.7
Children saying yes
P共x 20兲 P共x < 20.5兲 P共z < 3.58兲 ⬇ 0 54. n 12, p 0.33 → np 3.96 < 5 Cannot use normal distribution. P共x > 5兲 1 P共x 5兲 1 关P共x 0兲 P共x 1兲 . . . P共x 5兲兴 1 关12C0共0.33兲0共0.67兲12 12C1共0.33兲1共0.67兲11 12C2共0.33兲2共0.67兲10 12C3共0.33兲3共0.67兲9 12C4共0.33兲4共0.67兲8 12C5共0.33兲5共0.67兲7兴 1 0.8289 ⬇ 0.1711
CHAPTER 5 QUIZ SOLUTIONS 1. (a) P共z > 2.10兲 0.9821 (b) P共z < 3.22兲 0.9994 (c) P共2.33 < z < 2.33兲 0.9901 0.0099 0.9802 (d) P共z < 1.75 or z > 0.75兲 0.0401 0.7734 0.8135
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153
154
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2. (a) z z

NORMAL PROBABILITY DISTRIBUTIONS
x 5.36 5.5 ⬇ 1.75 0.08 x 5.64 5.5 ⬇ 1.75 0.08
P共5.36 < x < 5.64兲 P共1.75 < z < 1.75兲 0.9599 0.0401 0.9198 (b) z z
x 5.00 共8.2兲 ⬇ 0.41 7.84 x 0 共8.2兲 ⬇ 1.05 7.84
P共5.00 < x < 0兲 P共0.41 < z < 1.05兲 0.8531 0.6591 0.1940 (c) z z
x 0 18.5 2 9.25 x 37 18.5 2 9.25
P共x < 0 or x > 37兲 P共z < 2 or z > 2兲 2共0.0228兲 0.0456 3. z
x 320 290 ⬇ 0.81 37
P共x > 320兲 P共z > 0.81兲 0.2090 4. z z
x 250 290 ⬇ 1.08 37 x 300 290 ⬇ 0.27 37
P共250 < x < 300兲 P共1.08 < z < 0.27兲 0.6064 0.1401 0.4663 5. P共x > 250兲 P共z > 1.08兲 0.8599 → 85.99% 6. z
x 280 290 ⬇ 0.27 37
P共x < 280兲 P共z < 0.27兲 0.3936
共2000兲共0.3936兲 787.2 ⬇ 787 students 7. top 5% → z ⬇ 1.645
z 290 共1.645兲共37兲 350.9 ⬇ 351 8. bottom 25% → z ⬇ 0.67
z 290 共0.67兲共37兲 265.2 ⬇ 265 9. z
x 300 290 10 ⬇ 2.09 37 4.78 冪n 冪60
P共x > 300兲 P共z > 2.09兲 ⬇ 0.0183
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CHAPTER 5
10. z

NORMAL PROBABILITY DISTRIBUTIONS
x 300 290 ⬇ 0.27 37
P共x > 300兲 P共z > 0.27兲 0.3936 z
x 300 290 10 ⬇ 2.09 37 4.78 冪n 冪15
P共x > 300兲 P共z > 2.09兲 0.0183 You are more likely to select one student with a test score greater than 300 because the standard error of the mean is less than the standard deviation. 11. n 24, p 0.75 → np 18, nq 6 Use normal distribution.
np 18 12. z
冪npq ⬇ 2.121
x 15.5 18 ⬇ 1.18 2.12
P共x 15兲 P共x < 15.5兲 P共z < 1.18兲 0.1190
CUMULATIVE REVIEW, CHAPTERS 3 –5 1. (a) np 30共0.56兲 16.8 > 5 np 30共0.44兲 13.2 > 5 Use normal distribution. (b) np 30共0.56兲 16.8
冪npq 冪30共0.56兲共0.44兲 2.72 P共x 14兲 ⬇ P共x 14.5兲
冢
P z
14.5 16.8 2.72
冣
P共z 0.85兲 0.1977 (c) It is not unusual for 14 out of 30 employees to say they do not use all of their vacation time because the probability is greater than 0.05. P共x 14兲 ⬇ P共13.5 x 14.5兲 P
16.8 14.5 16.8 z 冢13.52.72 冣 2.72
P共1.21 z 0.85兲 0.1977 0.1131 0.0846
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155
156 2.
CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
x
P 冇 x冈
xP冇 x冈
xⴚ
冇x ⴚ 冈 2
冇x ⴚ 冈 2P冇x冈
2 3 4 5 6 7
0.421 0.233 0.202 0.093 0.033 0.017
0.842 0.699 0.808 0.465 0.198 0.119
1.131 0.131 0.869 1.869 2.869 3.869
1.279 0.017 0.755 3.493 8.231 14.969
0.539 0.004 0.153 0.325 0.272 0.254
兺xP共x兲 3.131
兺共x 兲 P共x兲 1.546 2
兺 xP共x兲 3.1 兺共x 兲 P共x兲 1.5
(a) (b) 2
2
(c) 冪 2 1.2 (d)
兺共x兲 3.1
(e) The expected family household size is 3.1 people with a standard deviation of 1.2 persons. 3.
x
P 冇 x冈
xP冇 x冈
xⴚ
冇x ⴚ 冈 2
冇x ⴚ 冈 2P冇x冈
0 1 2 3 4 5 6
0.012 0.049 0.159 0.256 0.244 0.195 0.085
0.000 0.049 0.318 0.768 0.976 0.975 0.510
3.596 2.596 1.596 0.596 0.404 1.404 2.404
12.931 6.739 2.547 0.355 0.163 1.971 5.779
0.155 0.330 0.405 0.091 0.040 0.384 0.491
兺xP共x兲 3.596 (a)
兺共x 兲 P共x兲 1.897 2
兺 xP共x兲 ⬇ 3.6
(b) 2
兺共x 兲 P共x兲 ⬇ 1.9 2
(c) 冪 2 ⬇ 1.4 (d)
兺共x兲 ⬇ 3.6
(e) The expected number of fouls per game is 3.6 with a standard deviation of 1.4 fouls. 4. (a) P共x < 4兲 0.012 0.049 0.159 0.256 0.476 (b) P共x 3兲 1 P共x 2兲 1 共0.012 0.049 0.159兲 0.78 (c) P共2 x 4兲 0.159 0.256 0.244 0.659
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CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
5. (a) 共16兲共15兲共14兲共13兲 43,680 (b)
840 共7兲共6兲共5兲共4兲 0.0192 共16兲共15兲共14兲共13兲 43,680
6. 0.9382
7. 0.0010
8. 1 0.2005 0.7995
9. 0.9990 0.500 0.4990 11. 0.5478 共1 0.9573兲 0.5905
10. 0.3974 0.1112 0.2862 12. n 10, p 0.78 (a) P共6兲 0.1108 (b) P共x 6兲 0.9521
(c) P共x < 6兲1 P共x 6兲 1 0.952 0.0479 13. p
1 0.005 200
(a) P共x 10兲 共0.005兲共0.995兲9 0.0048 (b) P共x 3兲 0.0149 (c) P共x > 10兲 1 P共x 10兲 1 0.0489 0.9511 14. (a) 0.224 (b) 0.879 (c) Dependent because a person can be a public school teacher and have more than 20 years experience. (d) 0.879 0.137 0.107 0.909 (e) 0.329 0.121 0.040 0.410 15. (a) x 70
x
冪n
1.2 冪40
0.1897
x 69.2
70
70.8
Initial pressure (in psi)
冢
(b) P共x 69兲 P z
69 70 1.2 冪15
冣
P共z < 3.23兲 0.0006
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157
158
CHAPTER 5

NORMAL PROBABILITY DISTRIBUTIONS
冢
16. (a) P共x < 36兲 P z <
36 44 P共z < 1.6兲 0.0548 5
(b) P共42 < x < 60兲 P
冣
冢42 5 44 < z < 60 5 44冣
P共0.40 < z < 3.2兲 0.9993 0.3446 0.6547 (c) Top 5% ⇒ z 1.645 x z 44 共1.645兲共5兲 52.2 months 17. (a) (b)
12C4
495
共1兲共1兲共1兲共1兲 0.0020 12C4
18. n 20, p 0.41 (a) P共8兲 0.1790 (b) P共x 6兲 1 P共x 5兲 1 0.108 0.8921 (c) P共x 13兲 1 P共x 14兲 1 0.0084 0.9916
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Confidence Intervals
CHAPTER
6
6.1 CONFIDENCE INTERVALS FOR THE MEAN (LARGE SAMPLES)
6.1 Try It Yourself Solutions 1a. x 14.8 b. The mean number of sentences per magazine advertisement is 14.8. 2a. Zc 1.96, n 30, s 16.5 b. E Zc
s n
1.96
16.5 5.9 30
c. You are 95% confident that the maximum error of the estimate is about 5.9 sentences per magazine advertisement. 3a. x 14.8, E 5.9 b. x E 14.8 5.9 8.9 x E 14.8 5.9 20.7 c. You are 95% confident that the mean number of sentences per magazine advertisements is between 8.9 and 20.7. 4b. 75% CI: 11.6, 13.2 85% CI: 11.4, 13.4 99% CI: 10.6, 14.2 c. The width of the interval increases as the level of confidence increases. 5a. n 30, x 22.9, 1.5, Zc 1.645 b. E Zc
1.5 1.645 0.451 0.5 n 30
x E 22.9 0.451 22.4 x E 22.9 0.451 23.4 c. You are 90% confident that the mean age of the students is between 22.4 and 23.4 years. 6a. Zc 1.96, E 2, s 5.0 b. n
E 1.962 5.0 Zc s
2
2
24.01 → 25
c. You should have at least 25 magazine advertisements in your sample.
6.1 EXERCISE SOLUTIONS 1. You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will equal the population mean exactly. 2. b
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160
CHAPTER 6

CONFIDENCE INTERVALS
3. d; As the level of confidence increases, zc increases therefore creating wider intervals. 4. b; As n increases, E decreases because n is in the denominator of the formula for E. Therefore, the intervals become narrower. 5. 1.28
6. 1.44
7. 1.15
8. 2.17
9. x 3.8 4.27 0.47
10. x 9.5 8.76 0.74
11. x 26.43 24.67 1.76
12. x 46.56 48.12 1.56
13. x 0.7 1.3 0.60
14. x 86.4 80.9 5.5
15. E zc 17. E zc
s n
1.645
2.5 0.685 36
16. E zc
0.93
1.5 0.197 50
18. E zc
s n
s n
s n
1.96 2.24
3.0 60
0.759
4.6 1.030 100
19. c 0.88 ⇒ zc 1.55 x 57.2, s 7.1, n 50 x ± zc
s n
57.2 ± 1.55
7.1 50
57.2 ± 1.556 55.6, 58.8
Answer: (c) 20. c 0.90 ⇒ zc 1.645 x 57.2, s 7.1, n 50 x ± zc
s n
57.2 ± 1.645
7.1 50
57.2 ± 1.652 55.5, 58.9
Answer: (d) 21. c 0.95 ⇒ zc 1.96 x 57.2, s 7.1, n 50 x ± zc
s n
57.2 ± 1.96
7.1 57.2 ± 1.968 55.2, 59.2 50
Answer: (b) 22. c 0.98 ⇒ zc 2.33 x 57.2, s 7.1, n 50 x ± zc
s n
57.2 ± 2.33
7.1 57.2 ± 2.340 54.9, 59.5 50
Answer: (a) 23. x ± zc 24. x ± zc
s n
s n
15.2 ± 1.645 31.39 ± 1.96
2.0 60
0.8 82
15.2 ± 0.425 14.8, 15.6 31.3 ± 0.173 31.22, 31.56
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CHAPTER 6
25. x ± zc 26. x ± zc
s n
s n
4.27 ± 1.96
0.3 42
13.5 ± 2.575

CONFIDENCE INTERVALS
4.27 ± 0.091 4.18, 4.36
1.5 100
13.5 ± 0.386 13.1, 13.9
27. (0.264, 0.494 ⇒ 0.379 ± 0.115 ⇒ x 0.379, E 0.115 28. (3.144, 3.176 ⇒ 3.16 ± 0.016 ⇒ x 3.16, E 0.016 29. (1.71, 2.05 ⇒ 1.88 ± 0.17 ⇒ x 1.88, E 0.17 30. (21.61, 30.15 ⇒ 25.88 ± 4.27 ⇒ x 25.88, E 4.27 31. c 0.90 ⇒ zc 1.645 n
zE 1.64516.8 c
2
2
125.13 ⇒ 126
32. c 0.95 ⇒ zc 1.96 n
zE 1.9612.5 c
2
2
24.01 ⇒ 25
33. c 0.80 ⇒ zc 1.28 n
zc E
2
1.284.1 2
2
6.89 ⇒ 7
34. c 0.98 ⇒ zc 2.33 n
zE 2.332 10.1 c
2
2
138.45 ⇒ 139
35. 2.1, 3.5 ⇒ 2E 3.5 2.1 1.4 ⇒ E 0.7 and x 2.1 E 2.1 0.7 2.8 36. 44.0780.97 ⇒ 2E 80.97 44.07 36.9 ⇒ E 18.45 and x 44.07 E 44.07 18.45 62.52 37. 90% CI: x ± zc 95% CI: x ± zc
s n
s n
630.90 ± 1.645
56.70 630.9 ± 16.49 614.41, 647.39 32
56.70 630.9 ± 19.65 611.25, 650.55 32
630.90 ± 1.96
The 95% confidence interval is wider. 38. 90% CI: x ± zc 95% CI: x ± zc
s n
s n
23.20 ± 1.645 23.20 ± 1.96
4.34 35
4.34 35
23.20 ± 1.207 21.99, 24.41
23.20 ± 1.438 21.76, 24.64
The 95% CI is wider.
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161
162
CHAPTER 6

39. 90% CI: x ± zc 95% CI: x ± zc
CONFIDENCE INTERVALS
s n
s n
99.3 ± 1.645 99.3 ± 1.96
41.5 31
99.3 ± 12.26 87.0, 111.6
41.5 99.3 ± 14.61 84.7, 113.9 31
The 95% confidence interval is wider. 40. 90% CI: x ± zc 95% CI: x ± zc
s n
s n
23 ± 1.645
6.7 23 ± 1.837 21, 25 36
23 ± 1.96
6.7 36
23 ± 2.189 21, 25
The 90% CI and the 95% CI have the same width. 41. x ± zc 42. x ± zc 43. x ± zc
s n
s n
s n
120 ± 1.96
17.50 120 ± 5.423 114.58, 125.42 40
150 ± 2.575 120 ± 1.96
15.5 150 ± 5.15 144.85, 155.15 60
17.50 120 ± 3.8348 116.17, 123.83 80
n 40 CI is wider because a smaller sample was taken giving less information about the population. 44. x ± zc
s n
150 ± 2.575
15.5 40
150 ± 6.31 143.69, 156.31
n 40 CI is wider because a smaller sample was taken giving less information about the population. 45. x ± zc 46. x ± zc 47. x ± zc
s n
s n
s n
3.12 ± 2.575
0.09 3.12 ± 0.033 3.09, 3.15 48
107.05 ± 2.575 3.12 ± 2.575
28.10 107.05 ± 9.26 97.79, 116.31 61
0.06 48
3.12 ± 0.022 3.10, 3.14
s 0.09 CI is wider because of the increased variability within the sample. 48. x ± zc
s n
107.05 ± 2.575
32.50 107.05 ± 10.72 96.33, 117.77 61
s 32.10 CI is wider because of the increased variability within the population. 49. (a) An increase in the level of confidence will widen the confidence interval. (b) An increase in the sample size will narrow the confidence interval. (c) An increase in the standard deviation will widen the confidence interval. 50. Answers will vary.
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CHAPTER 6
51. x

CONFIDENCE INTERVALS
x 136 9.1 n 15
90% CI: x ± zc 99% CI: x ± zc
n
9.1 ± 1.645
1.5 15
9.1 ± 0.637 8.4, 9.7
1.5 9.1 ± 2.575 9.1 ± 0.997 8.1, 10.1 n 15
99% CI is wider. 52. 90% CI: x ± zc 99% CI: x ± zc
n
1.7 ± 1.645
0.6 21
1.7 ± 0.215 1.5, 1.9
0.6 1.7 ± 2.575 1.7 ± 0.337 1.4, 2.0 n 21
99% CI is wider. 53. n
E 1.961 4.8
54. n
E 2.5752 1.4
zc zc
55. (a) n (b) n
2
2
2
88.510 → 89 2
3.249 → 4
E 1.960.5 2.8 zc
2
2
120.473 → 121 servings
E 2.5750.5 2.8 zc
2
2
207.936 → 208 servings
99% CI requires larger sample because more information is needed from the population to be 99% confident. 56. (a) n
E 1.6451 1.2
(b) n
E 2.5751 1.2
zc
zc
2
2
2
3.897 → 4 students
2
9.548 → 10 students
99% CI requires larger sample because more information is needed from the population to be 99% confident. 57. (a) n
E 1.6450.25 0.85
(b) n
E 1.6450.15 0.85
zc
zc
2
2
2
2
31.282 → 32 cans 86.893 → 87 cans
E 0.15 requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain enough information from the population to ensure desired accuracy. 58. (a) n
E 1.961 3
(b) n
E 1.962 3
zc
zc
2
2
2
2
34.574 → 35 bottles 8.644 → 9 bottles
E 1 requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain enough information from the population to ensure desired accuracy.
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163
164
CHAPTER 6

CONFIDENCE INTERVALS
59. n
0.25 E 1.9600.125
2
n
0.25 E 1.960 0.0625
2
zc
zc
2
2
15.3664 → 16 sheets 61.4656 → 62 sheets
E 0.0625 requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain enough information from the population to ensure desired accuracy. 60. (a) n
0.15 E 1.645 0.0425
(b) n
0.15 E 1.645 0.02125
zc
zc
2
2
2
2
33.708 → 34 units 134.833 → 135 units
E 0.02125 requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain enough information from the population to ensure desired accuracy. 61. (a) n
E 2.5750.1 0.25
(b) n
E 2.5750.1 0.30
zc
zc
2
2
2
2
41.441 → 42 soccer balls 59.676 → 60 soccer balls
0.30 requires a larger sample size. Due to the increased variability in the population, a larger sample size is needed to ensure the desired accuracy. 62. (a) n
E 2.5750.15 0.20
(b) n
E 2.5750.15 0.10
zc
zc
2
2
2
2
11.788 → 12 soccer balls 2.947 → 3 soccer balls
0.20 requires a larger sample size. Due to the decreased variability in the population, a smaller sample is needed to ensure desired accuracy. 63. (a) An increase in the level of confidence will increase the minimum sample size required. (b) An increase (larger E) in the error tolerance will decrease the minimum sample size required. (c) An increase in the population standard deviation will increase the minimum sample size required. 64. A 99% CI may not be practical to use in all situations. It may produce a CI so wide that it has no practical application. 65. x 238.77, s 13.20, n 31 s 13.20 x ± zc 238.77 ± 1.96 238.77 ± 4.65 234.1, 243.4 n 31 66. x 99.08, s 12.22, n 37 x ± zc
s n
99.08 ± 1.96
12.22 99.08 ± 3.94 95.1, 103.0 37
67. x 15.783, s 2.464, n 30 s 2.464 x ± zc 15.783 ± 1.96 15.783 ± 0.88173 14.902, 16.665 n 30 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 6

CONFIDENCE INTERVALS
68. x 16.656, s 4.235, n 33 x ± zc 69. (a) (c)
s n
16.656 ± 1.96
4.235 16.656 ± 1.445 15.211, 18.101 33
500 0.707 NN n1 1000 1000 1
(b)
100 0.949 NN n1 1000 1000 1
75 0.962 NN n1 1000 1000 1
(d)
50 0.975 NN n1 1000 1000 1
(e) The finite population correction factor approaches 1 as the sample size decreases while the population size remains the same. 70. (a)
50 0.711 NN n1 100 100 1
(b)
50 0.937 NN n1 400 400 1
(c)
50 0.964 NN n1 700 700 1
(d)
50 0.988 NN n1 2000 2000 1
(e) The finite population correction factor approaches 1 as the population size increases while the sample size remains the same. 71. E
zc z z → n c → n c n E E
2
6.2 CONFIDENCE INTERVALS FOR THE MEAN (SMALL SAMPLES)
6.2 Try It Yourself Solutions 1a. d.f. n 1 22 1 21 b. c 0.90 c. 1.721 2a. 90% CI: tc 1.753 E tc
s n
1.753
10 4.383 4.4 16
99% CI: tc 2.947 s 10 E tc 2.947 7.368 7.4 n 16 b. 90% CI: x ± E 162 ± 4.383 157.6, 166.4 99% CI: x ± E 162 ± 7.368 154.6, 169.4 c. You are 90% confident that the mean temperature of coffee sold is between 157.6° and 166.4°. You are 99% confident that the mean temperature of coffee sold is between 154.6° and 169.4°.
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165
166

CHAPTER 6
CONFIDENCE INTERVALS
3a. 90% CI: tc 1.729 s 0.42 E tc 1.729 0.162 0.16 n 20 95% CI: tc 2.093 s 0.42 E tc 2.093 0.197 0.20 n 20 b. 90% CI: x ± E 6.22 ± 0.162 6.06, 6.38 95% CI: x ± E 6.22 ± 0.197 6.02, 6.42 c. You are 90% confident that the mean mortgage interest rate is contained between 6.06% and 6.38%. You are 95% confident that the mean mortgage interest rate is contained between 6.02% and 6.42%. 4a. Is n 30? No Is the population normally distributed? Yes Is known? No Use the tdistribution to construct the 90% CI.
6.2 EXERCISE SOLUTIONS 1. 1.833 5. E tc 7. E tc
2. 2.201 s
2.131
n
s
1.796
n
9. (a) x ± tc (b) x ± tc
s s
5
4. 2.539
2.664 2.7
6. E tc
2.4 1.244 1.2 12
8. E tc
16
12.5 ± 2.015
n
n
3. 2.947
12.5 ± 1.645
s n
s n
4.032 2.896
3 6
4.938 4.9
4.7 4.537 4.5 9
2.0 12.5 ± 1.645 10.9, 14.1 6 2.0 6
12.5 ± 1.343 11.2, 13.8
tCI is wider. 10. (a) x ± tc
s n
(b) x ± zc
s n
13.4 ± 2.365
0.85 13.4 ± 0.711 12.7, 14.1 8
13.4 ± 1.96
0.85 13.4 ± 0.589 12.8, 14.0 8
tCI is wider. 11. (a) x ± tc
s n
(b) x ± zc
4.3 ± 2.650
s n
0.34 4.3 ± 0.241 4.1, 4.5 14
4.3 ± 2.326
0.34 14
4.3 ± 0.211 4.1, 4.5
Both CIs have the same width.
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CHAPTER 6
s
12. (a) x ± tc
s
(b) x ± zc
4.6
24.7 ± 3.250
n
n
CONFIDENCE INTERVALS
24.7 ± 4.728 20.0, 29.4
10
24.7 ± 2.575

4.6 10
24.7 ± 3.746 21.0, 28.4
tCI is wider. 13. x ± tc
s
75 ± 2.776
n
E tc
s s
14. x ± tc
15. x ± zc
100 ± 2.447
n
s
E tc
2.447
n
1.96
n
42.50 100 ± 39.307 60.69, 139.31 7
42.50 39.307 39.31 7
75 ± 1.96
n
E zc
12.50 15.518 5
2.776
n
12.50 75 ± 15.518 59.48, 90.52 5
15 5
15 5
75 ± 13.148 61.85, 88.15
13.148 13.15
tCI is wider. 16. x ± tc
50 100 ± 1.96 100 ± 37.041 62.96, 137.04 n 7
E zc
n
1.96
50 7
37.041 37.04
tCI is wider. 17. (a) x ± tc
s n
(b) x ± zc
4.54 ± 1.833
s
1.21 10
1.21
4.54 ± 1.645
n
4.54 ± 0.701 3.84, 5.24
500
4.54 ± 0.089 4.45, 4.63
tCI is wider. 18. (a) x ± tc
s
1.46 ± 1.796
n
(b) x ± zc
s n
0.28 1.46 ± 0.145 1.31, 1.61 12
1.46 ± 1.645
0.28 1.46 ± 0.019 1.44, 1.48 600
tCI is wider. 19. (a) x 4460.16 (c) x ± tc
s n
4460.16 ± 3.250
20. (a) x 5627.652 (c) x ± tc
s n
(b) s 146.143 146.143 4460.16 ± 150.197 4309.96, 4610.36 10
(b) s 336.186
5627.652 ± 3.012
336.186 5627.652 ± 270.627 5357.025, 5898.279 14
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167
168
CHAPTER 6

CONFIDENCE INTERVALS
21. (a) x 1767.7 (b) s 252.23 252.2 (c) x ± tc
s n
1767.7 ± 3.106
252.23 1767.7 ± 226.16 1541.5, 1993.8 12
22. (a) x 2.35 (b) s 1.03 (c) x ± tc
s n
2.35 ± 2.977
1.03 2.35 ± 0.793 1.56, 3.14 15
23. n 30 → use normal distribution x ± zc
s n
1.25 ± 1.96
0.05 1.25 ± 0.012 1.24, 1.26 70
24. x 57.79, s 19.05, n < 30, unknown, and pop normally distributed → use tdistribution x ± tc
s n
57.79 ± 2.201
19.05 57.79 ± 12.104 45.69, 69.89 12
25. x 21.9, s 3.46, n < 30, known, and pop normally distributed → use tdistribution s 3.46 x ± tc 21.9 ± 2.064 21.9 ± 1.43 20.5, 23.3 n 25 26. x 20.8, s 4.5, n < 30, known, and pop normally distributed → use normal distribution 4.7 x ± zc 20.8 ± 1.96 20.8 ± 2.06 18.7, 22.9 n 20 27. n < 30, unknown, and pop not normally distributed → cannot use either the normal or tdistributions. 28. n < 30, unknown, and pop normally distributed → use tdistribution s 61.32 x ± tc 144.19 ± 2.110 144.19 ± 30.496 113.69, 174.69 n 18 29. n 25, x 56.0, s 0.25 ± t0.99 → 99% tCI
x ± tc
s n
56.0 ± 2.797
0.25 25
56.0 ± 0.140 55.9, 56.1
They are not making good tennis balls because desired bounce height of 55.5 inches is not contained between 55.9 and 56.1 inches. 30. n 16, x 1015, s 25 ± t 0.99 → 99% tCI
x ± tc
s n
1015 ± 2.947
25 16
1015 ± 18.419 997, 1033
They are making good light bulbs because the desired bulb life of 1000 hours is contained between 997 and 1033 hours.
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CHAPTER 6

CONFIDENCE INTERVALS
6.3 CONFIDENCE INTERVALS FOR POPULATION PROPORTIONS
6.3 Try It Yourself Solutions 1a. x 181, n 1006 b. p ^
181 0.180 1006
2a. p 0.180, q 0.820 ^
^
b. np 10060.180 181.08 > 5 ^
nq 10060.820 824.92 > 5 ^
c. zc 1.645
0.820 0.020 pnq 1.6450.1801006 ^ ^
E zc
d. p ± E 0.180 ± 0.020 0.160, 0.200 ^
e. You are 90% confident that the proportion of adults say that Abraham Lincoln was the greatest president is contained between 16.0% and 20.0%. 3a. n 900, p 0.33 ^
b. q 1 p 1 0.33 0.67 ^
^
c. np 900
0.33 297
> 5
nq 900
0.67 603
> 5
^
^
^
Distribution of p is approximately normal. d. zc 2.575
e. p ± zc ^
0.33900 0.67 0.33 ± 0.04 0.290, 0.370
pq 0.33 ± 2.575 n ^ ^
f. You are 99% confident that the proportion of adults who think that people over 75 are more dangerous drivers is contained between 29.0% and 37.0%. 4a. (1) p 0.50, q 0.50 zc 1.645, E 0.02 ^
^
(2) p 0.064, q 0.936 ^
^
zc 1.645, E 0.02 b. (1) n pq ^ ^
(2) n pq ^ ^
zE c
E zc
2
2
0.500.50 0.064
1.645 0.02
2
0.936 0.02 1.645
1691.266 → 1692 2
405.25 → 406
c. (1) At least 1692 males should be included in the sample. (2) At least 406 males should be included in the sample.
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169
170
CHAPTER 6

CONFIDENCE INTERVALS
6.3 EXERCISE SOLUTIONS 1. False. To estimate the value of p, the population proportion of successes, use the point x estimate p . n ^
2. True 3. p ^
x 752 0.750 n 1002
4. p ^
q 1 p 0.250 ^
5. p ^
q 1 p 0.170
^
^
x 2938 0.663 n 4431
6. p ^
q 1 p 0.337 ^
7. p ^
9. p ^
^
x 144 0.170 n 848
8. p ^
11. p ^
x 224 0.489 n 458 ^
x 110 0.110 n 1003
q 1 p 0.890
^
^
x 204 0.718 n 284
10. p ^
q 1 p 0.282 ^
^
q 1 p 0.511
^
q 1 p 0.830 ^
x 2439 0.830 n 2939
^
x 662 0.660 n 1003
q 1 p 0.340
^
^
x 230 0.230 n 1000
12. p ^
q 1 p 1 0.230 0.770
^
x 1515 0.470 n 3224
q 1 p 1 0.470 0.530
^
^
13. p 0.48, E 0.03
^
14. p 0.15, E 0.052
^
^
p ± E 0.48 ± 0.03 0.45, 0.51
p ± E 0.51 ± 0.052 0.458, 0.562
^
^
Note: Exercises 15–20 may have slightly different answers from the text due to rounding.
15. 95% CI: p ± zc ^
^ ^
99% CI: p ± zc ^
0.7501002 0.25 0.750 ± 0.027 0.723, 0.777
pq 0.750 ± 1.96 n
0.250 0.75 ± 0.035 0.715, 0.785 0.7501002
pq 0.750 ± 2.575 n ^ ^
99% CI is wider.
0.170 0.830 ± 0.014 0.816, 0.845 0.8302939 pq 0.830 0.170 0.830 ± 2.575 0.830 ± 0.018 0.812, 0.848 99% CI: p ± z n 2939 pq 0.830 ± 1.96 n ^ ^
16. 95% CI: p ± zc ^
^ ^
^
c
99% CI is wider.
17. 95% CI: p ± zc ^
99% CI: p ± zc ^
0.337 0.663 ± 0.014 0.649, 0.677 0.6634431
pq 0.663 ± 1.96 n ^ ^
0.337 0.663 ± 0.018 0.645, 0.681 0.6634431
pq 0.663 ± 2.575 n ^ ^
99% CI is wider. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 6

CONFIDENCE INTERVALS
pnq 0.489 ± 1.960.489458 0.511 0.489 ± 0.046 0.433, 0.535 ^ ^
18. 95% CI: p ± zc ^
pnq 0.489 ± 2.5750.489458 0.511 0.489 ± 0.060 0.429, 0.549 ^ ^
99% CI: p ± zc ^
99% CI is wider.
19. 95% CI: p ± zc ^
^ ^
99% CI: p ± zc ^
0.170848 0.830 0.170 ± 0.025 0.145, 0.195
pq 0.170 ± 1.96 n
0.170848 0.830 0.170 ± 0.033 0.137, 0.203
pq 0.170 ± 2.575 n ^ ^
99% CI is wider.
0.890 0.110 ± 0.019 0.091, 0.129 0.1101003 pq 0.110 0.890 0.110 ± 2.575 0.110 ± 0.025 0.085, 0.135 99% CI: p ± z n 1003 pq 0.110 ± 1.96 n ^ ^
20. 95% CI: p ± zc ^
^ ^
^
c
99% CI is wider. 21. (a) n pq ^ ^
E zc
(b) n pq ^ ^
2
E zc
2
0.5
0.5 0.03 1.96
0.26
2
1067.111 → 1068 vacationers
0.74 0.03 1.96
2
821.249 → 822 vacationers
(c) Having an estimate of the proportion reduces the minimum sample size needed. 22. (a) n pq ^ ^
zc E
(b) n pq ^ ^
2
E zc
2
0.5
0.5 0.04
0.3
0.7 0.04
2.33
2.33
2
2
848.27 → 849 vacationers 712.54 → 713 vacationers
(c) Having an estimate of the proportion reduces the minimum sample size needed. 23. (a) n pq ^ ^
E zc
(b) n pq ^ ^
2
E zc
2
0.5
0.5 0.025 2.05
0.25
2
1681 camcorders
0.75 0.025 2.05
2
1260.75 → 1261 camcorders
(c) Having an estimate of the proportion reduces the minimum sample size needed. 24. (a) n pq ^ ^
(b) n pq ^ ^
zc E
2
E zc
2
0.5
0.5 0.035
0.19
2.17
2
961 computers
0.81 0.035 2.17
2
591.592 → 592 computers
(c) Having an estimate of the proportion reduces the minimum sample size needed.
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171
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CHAPTER 6

CONFIDENCE INTERVALS
25. (a) p 0.36, n 400 ^
p ± zc ^
0.36400 0.64 0.36 ± 0.062 0.298, 0.422
pq 0.36 ± 2.575 n ^ ^
(b) p 0.32, n 400 ^
p ± zc ^
0.32400 0.68 0.32 ± 0.060 0.260, 0.380
pq 0.32 ± 2.575 n ^ ^
It is possible that the two proportions are equal because the confidence intervals estimating the proportions overlap. 26. (a) p 0.26, n 400 ^
p ± zc ^
0.26400 0.74 0.26 ± 0.056 0.204, 0.316
pq 0.26 ± 2.575 n ^ ^
(b) p 0.56, n 400 ^
p ± zc ^
0.56400 0.44 0.56 ± 0.064 0.496, 0.624
pq 0.56 ± 2.575 n ^ ^
It is unlikely that the two proportions are equal because the confidence intervals estimating the proportions do not overlap. 27. (a) p 0.65, q 0.35, n 2563 ^
^
0.35 0.65 ± 0.024 0.626, 0.674 pnq 0.65 ± 2.5750.652563 ^ ^
p ± zc ^
(b) p 0.88, q 0.12, n 1125 ^
^
p ± zc ^
0.12 0.88 ± 0.025 0.855, 0.905 0.881125
pq 0.88 ± 2.575 n ^ ^
(c) p 0.92, q 0.08, n 1086 ^
^
p ± zc ^
0.08 0.92 ± 0.021 0.899, 0.941 0.921086
pq 0.92 ± 2.575 n ^ ^
28. (a) The two proportions are possibly unequal because the 2 CI’s 0.626, 0.674 and 0.855, 0.905 do not overlap. (b) The two proportions are possibly equal because the 2 CI’s 0.855, 0.905 and 0.899, 0.941 overlap. (c) The two proportions are possibly unequal because the 2 CI’s 0.626, 0.674 and 0.899, 0.941 do not overlap. 29. 31.4% ± 1% → 30.4%, 32.4% → 0.304, 0.324
E zc
pq → zc E n ^ ^
1.981 → z 1.98 → c 0.952 pnq 0.010.3148451 0.686 ^ ^
c
30.4%, 32.4% is approximately a 95.2% CI.
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CHAPTER 6

CONFIDENCE INTERVALS
30. 27% ± 3% → 24%, 30% → 0.24, 0.30 2.138 → z 2.14 → c 0.968 pnq → z Epnq 0.030.271001 0.73 ^ ^
E zc
c
c
^ ^
24%, 30% is approximately a 96.8% CI. 31. If np < 5 or nq < 5, the sampling distribution of p may not be normally distributed, therefore preventing the use of zc when calculating the confidence interval. ^
^
32. E zc 33.
^
pq E → n zc ^ ^
pq E → n zc ^ ^
2
pq z → n pq c n E ^ ^
^ ^
^ p
^ ⴝ1ⴚ^ q p
^q ^ p
^ p
^ ⴝ1ⴚ^ q p
^q ^ p
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
0.00 0.09 0.16 0.21 0.24 0.25 0.24 0.21 0.16 0.09 0.00
0.45 0.46 0.47 0.48 0.49 0.50 0.51 0.52 0.53 0.54 0.55
0.55 0.54 0.53 0.52 0.51 0.50 0.49 0.48 0.47 0.46 0.45
0.2475 0.2484 0.2491 0.2496 0.2499 0.2500 0.2499 0.2496 0.2491 0.2484 0.2475
2
p 0.5 give the maximum value of pq. ^
^ ^
6.4 CONFIDENCE INTERVALS FOR VARIANCE AND STANDARD DEVIATION 6.4 Try It Yourself Solutions 1a. d.f. n 1 24 level of confidence 0.95 b. Area to the right of R2 is 0.025. Area to the left of L2 is 0.975. c. R2 39.364, L2 12.401 2a. 90% CI: R2 42.557, L2 17.708 95% CI: R2 45.722, L2 16.047 b. 90% CI for 2:
1.2 , 29 1.2 0.98, 2.36 n 1s , n 1s 2942.557 17.708
95% CI for 2:
1.2 , 29 1.2 0.91, 2.60 n 1s , n 1s 2945.722 16.047
2
2 R
2
2
2 R
2
2
2 L
2
2
2
2 L
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173
174

CHAPTER 6
CONFIDENCE INTERVALS
c. 90% CI for : 0.981, 2.358 0.99, 1.54 95% CI for : 0.913, 2.602 0.96, 1.61 d. You are 90% confident that the population variance is between 0.98 and 2.36, and that the population standard deviation is between 0.99 and 1.54. You are 95% confident that the population variance is between 0.91 and 2.60, and that the population standard deviation is between 0.96 and 1.61.
6.4 EXERCISE SOLUTIONS 1. R2 16.919, L2 3.325
2. R2 28.299, L2 3.074
3. R2 35.479, L2 10.283
4. R2 44.314, L2 11.524
5. R2 52.336, L2 13.121
6. R2 37.916, L2 18.939
7. (a) s 0.00843
0.00843 13 0.00843 , n 1s , n 1s 13 22.362 0.0000413, 0.000157 5.892 2
2
2
2
2 L
2 R
(b) 0.0000413, 0.000157 0.00643, 0.0125 8. (a) s 0.0321
0.0321 14 0.0321 , n 1s , n 1s 14 23.685 0.000610, 0.00220 6.571 2
2 R
2
2
2
2 L
(b) 0.000609, 0.00220 0.0247, 0.0469 9. (a) s 0.253
0.253 , 17 0.253 0.0305, 0.191 n 1s , n 1s 17 35.718 5.697 2
2 R
2
2
2
2 L
(b) 0.0305, 0.191 0.175, 0.437 10. (a) s 0.0918
0.0918 16 0.0918 , n 1s , n 1s 16 28.845 0.00467, 0.0195 6.908 2
2 R
2
2
2
2 L
(b) 0.00467, 0.0195 0.0683, 0.140
3.25 , 11 3.25 4.34, 44.64 n 1s , n 1s 1126.757 2.603 2
11. (a)
2 R
2
2
2
2 L
(b) 4.342, 44.636 2.08, 6.68
123 , 13 123 7951, 39,265 n 1s , n 1s 1324.736 5.009 2
12. (a)
2 R
2
2
2
2 L
(b) 7951, 39,265 89, 198
26 , 9 26 359.6, 1829.8 n 1s , n 1s 916.919 3.325 2
13. (a)
2 R
2
2
2
2 L
(b) 359.596, 1829.774 19.0, 42.8
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CHAPTER 6
14. (a)

CONFIDENCE INTERVALS
n 1s2 n 1s2 156.422 156.422 , , 22.5, 98.7 R2 L2 27.488 6.262
(b) 22.5, 98.7 4.7, 9.9
15 , 18 15 128, 492 n 1s , n 1s 1831.526 8.231 2
15. (a)
2 R
2
2
2
2 L
(b) 128.465, 492.042 11, 22
3600 293600 , 8,831,450, 21,224,305 n 1s , n 1s 2942.557 17.708 2
16. (a)
2 R
2
2
2
2 L
(b) 8,831,450, 21,224,305 2972, 4607
342 , 13 342 61,470, 303,559 n 1s , n 1s 1324.736 5.009 2
17. (a)
2 R
2
2
2
2 L
(b) 61,470.41, 303,559.99 248, 551 18. (a)
n 1s2 n 1s2 29 2.462 29 2.462 , , 3.35, 13.38 R2 L2 52.336 13.121
(b) 3.35, 13.38 1.83, 3.66
3.6 213.6 , 7.0, 30.6 n 1s , n 1s 2138.932 8.897 2
19. (a)
2 R
2
2
2
2 L
(b) 6.99, 30.59 2.6, 5.5
3900 193900 , 9,586,982, 28,564,792 n 1s , n 1s 1930.144 10.117 2
20. (a)
2 R
2
2
2
2 L
(b) 9,586,982, 28,564,792 3096, 5345 21. 90% CI for : 0.00643, 0.0125 Yes, because the confidence interval is below 0.015. 22. 90% CI for : 0.0247, 0.0469 No, because the majority of the confidence interval is above 0.025.
CHAPTER 6 REVIEW EXERCISE SOLUTIONS 1. (a) x 103.5 (b) s 34.663 s 34.663 E zc 1.645 9.0 n 40 2. (a) x 9.5 (b) s 7.1 E zc 3. x ± zc
s n
s n
1.645
7.1 2.1 32
10.3 ± 1.96
0.277 100
10.3 ± 0.054 10.2, 10.4
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175
176
CHAPTER 6
s
4. x ± zc

CONFIDENCE INTERVALS
0.0013 0.0925 ± 0.0003 0.0922, 0.0928 45
0.0925 ± 1.645
n
5. s 34.663 n
E 1.96 1034.663 zc
2
2
46.158 ⇒ 47 people
6. n
E 2.575 234.663
7. n
z 1.9627.098
8. n
7.098 E 2.330.5
zc
2
2
c
zc
2
9. tc 2.365 n
s
15. E tc 17. x ± zc 18. x ± tc 19. x ± tc 20. x ± tc 21. x ± tc 22. x ± tc
n
s s
2.718
25.6
48.39 ⇒ 49 people
2
1094.06 ⇒ 1095 people
s s
0.912 0.7
16. E tc
1.1
s
s n
1.1
2.064 2.861
25
0.5
16.520 10.6
72.1 ± 11.219 60.9, 83.3
3.5 ± 0.454 3, 4
6.520 25.2 ± 10.556 14.6, 35.8 80 ± 6.366 74, 86
15
80 ± 2.977
s n
0.912 6.8 ± 0.706 6.1, 7.5 14
80 ± 1.761
n
25.6 16
25
25.2 ± 2.861
n
12. tc 2.756 14. E tc
6.8 ± 2.718
n
11. tc 2.624
11.2
16
3.5 ± 2.064
n
s
1.753
72.1 ± 1.753
n
n
1991.713 ⇒ 1992 people
10. tc 1.721
s
13. E t c
2
2
14 15
80 ± 10.761 69, 91
23. p
x 560 0.28, q 0.72 n 2000
24. p
x 425 0.85, q 0.15 n 500
25. p
x 442 0.220, q 0.780 n 2010
26. p
x 90 0.113, q 0.887 n 800
27. p
116 x 0.180, q 0.820 n 644
28. p
594 x 0.590, q 0.410 n 1007
29. p
x 2021 0.420, q 0.580 n 4813
30. p
x 1230 0.520, q 0.480 n 2365
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
0.72 0.28 ± 0.020 0.260, 0.300 pnq 0.28 ± 1.960.282000 ^ ^
31. p ± zc ^
32. p ± zc ^
0.85500 0.15 0.85 ± 0.041 0.809, 0.891
pq 0.85 ± 2.575 n ^ ^
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CHAPTER 6
33. p ± zc ^

CONFIDENCE INTERVALS
0.780 0.220 ± 0.015 0.205, 0.235 0.2202010
pq 0.220 ± 1.645 n ^ ^
pnq 0.113 ± 2.3260.113800 0.887 0.113 ± 0.026 0.087, 0.139 ^ ^
34. p ± zc ^
^
pq 0.590 ± 1.645 n
pq 0.420 ± 1.96 n
36. p ± zc ^
37. p ± zc ^
0.180644 0.820 0.180 ± 0.039 0.141, 0.219
pq 0.180 ± 2.575 n
35. p ± zc
^ ^
0.410 0.590 ± 0.025 0.565, 0.615 0.5901007
^ ^
0.580 0.420 ± 0.014 0.406, 0.434 0.4204813
^ ^
0.48 0.52 ± 0.024 0.496, 0.544 pnq 0.52 ± 2.3260.522365 ^ ^
38. p ± zc ^
^ ^
(b) n pq ^ ^
E zc
2
E
2
39. (a) n pq
zc
0.50
0.50 0.05
0.63
0.37 0.05
1.96
1.96
2
2
384.16 → 385 adults 358.19 → 359 adults
(c) The minimum sample size needed is smaller when a preliminary estimate is available. 40. n pq ^ ^
zc E
2
0.63
0.37 0.025 2.575
2
2472.96 → 2473 adults
The sample size is larger. 41. R2 23.337, L2 4.404
42. R2 42.980, L2 10.856
43. R2 14.067, L2 2.167
44. R2 23.589, 2L 1.735
45. s 0.0727 95% CI for 2 :
0.0727 15 0.0727 , n 1s , n 1s 15 27.488 0.0029, 0.0127 6.262 2
2 R
2
2
2
2 L
95% CI for : 0.00288, 0.01266 0.0537, 0.1125 46. 99% CI for 2:
n 1s2 n 1s2 15 0.07272 15 0.07272 , 0.0024, 0.0172 , R2 L2 32.801 4.601
99% CI for : 0.00242, 0.01723 (0.0492, 0.1313 47. s 1.125 90% CI for 2 :
1.125 231.125 , 0.83, 2.22 n 1s , n 1s 2335.172 13.091 2
2 R
2
2
2
2 L
90% CI for : 0.83, 2.22 0.91, 1.49 48. 95% CI for 2:
n 1s2 n 1s2 231.1252 231.1252 , , 0.76, 2.49 R2 L2 38.076 11.689
95% CI for : 0.76, 2.49 (0.87, 0.158
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177
178
CHAPTER 6

CONFIDENCE INTERVALS
CHAPTER 6 QUIZ SOLUTIONS 1. (a) x 98.110 (b) s 24.722 s
E tc
1.960
n
s
(c) x ± tc
n
24.722 8.847 30
98.110 ± 1.960
24.722 98.110 ± 8.847 89.263, 106.957 30
You are 95% confident that the population mean repair costs is contained between $89.26 and $106.96. 2. n
2.57510 22.50 33.568 → 34 dishwashers zc
2
3. (a) x 12.96 (b) s 1.35 (c) x ± tc
s n
(d) x ± zc
12.96 ± 1.833
n
12.96 ± 1.645
1.35 10
2.63 10
12.96 ± 0.783 12.18, 13.74 12.96 ± 1.368 11.59, 14.33
The zCI is wider because 2.63 while s 1.35. 4. x ± tc
s n
5. (a) p ^
6824 ± 2.447
x 643 0.620 n 1037
(b) p ± zc ^
340 6824 ± 314.46 6510, 7138 7
(c) n pq ^ ^
^ ^
E zc
0.6201037 0.38 0.620 ± 0.025 0.595, 0.645
pq 0.620 ± 1.645 n 2
0.620
0.38 0.04 2.575
2
976.36 → 977 adults
Note: The answer for Exercise 6 may differ slightly from the text answer due to rounding.
24.722 29 24.722 , n 1s , n 1s 29 45.722 387.650, 1104.514 16.047 2
6. (a)
2 R
2
2
2
2 L
(b) 387.650, 1104.514 19.689, 33.234
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Hypothesis Testing with One Sample
CHAPTER
7
7.1 INTRODUCTION TO HYPOTHESIS TESTING
7.1 Try It Yourself Solutions 1a. (1) The mean . . . is not 74 months. ⫽ 74 (2) The variance . . . is less than or equal to 3.5. 2 ⱕ 3.5 (3) The proportion . . . is greater than 39%. p > 0.39 b. (1) ⫽ 74
(2) 2 > 3.5
(3) p ⱕ 0.39
c. (1) H0 : ⫽ 74; Ha : ⫽ 74; (claim) (2) H0 : 2 ⱕ 3.5 (claim); Ha : 2 > 3.5 (3) H0 : p ⱕ 0.39; Ha : p > 0.39 (claim) 2a. H0 : p ⱕ 0.01; Ha : p > 0.01 b. Type I error will occur if the actual proportion is less than or equal to 0.01, but you reject H0 . Type II error will occur if the actual proportion is greater than 0.01, but you fail to reject H0 . c. Type II error is more serious because you would be misleading the consumer, possibly causing serious injury or death. 3a. (1) H0 : ⫽ 74; Ha : ⫽ 74 (2) H0 : p ⱕ 0.39; Ha : p > 0.39 b. (1) Twotailed c. (1)
(2) Righttailed
1 Pvalue 2 area
−z
1 Pvalue 2 area
(2)
z z
Pvalue area z z
4a. There is enough evidence to support the radio station’s claim. b. There is not enough evidence to support the radio station’s claim. 5a. (1) Support claim.
(2) Reject claim.
b. (1) H0: ⱖ 650; Ha: < 650 (claim) (2) H0: ⫽ 98.6 (claim); Ha: ⫽ 98.6
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180

CHAPTER 7
HYPOTHESIS TESTING WITH ONE SAMPLE
7.1 EXERCISE SOLUTIONS 1. Null hypothesis 共H0兲 and alternative hypothesis 共Ha兲. One represents the claim, the other, its complement. 2. Type I Error: The null hypothesis is rejected when it is true. Type II Error: The null hypothesis is not rejected when it is false. 3. False. In a hypothesis test, you assume the null hypothesis is true. 4. False. A statistical hypothesis is a statement about a population. 5. True 6. True 7. False. A small Pvalue in a test will favor a rejection of the null hypothesis. 8. False. If you want to support a claim, write it as your alternative hypothesis. 9. H0 : ⱕ 645 (claim); Ha : > 645
10. H0 : ⱖ 128; Ha : < 128 (claim)
11. H0 : ⫽ 5; Ha : ⫽ 5 (claim)
12. H0 : 2 ⱖ 1.2 (claim); Ha : 2 < 1.2
13. H0 : p ⱖ 0.45; Ha: p < 0.45 (claim)
14. H0 : p ⫽ 0.21; Ha : p ⫽ 0.21
H0 : ⱕ 3
15. c,
16. d, Ha : ⱖ 3 μ
1
2
3
μ
4
1
H0 : ⫽ 3
17. b,
2
3
4
18. a, Ha: ⱕ 2 μ
1
2
3
19. Righttailed
μ
4
1
20. Lefttailed
2
3
21. Twotailed
4
22. Twotailed
23. > 750 H0: ⱕ 750; Ha: > 750 (claim)
24. < 3 H0 : ⱖ 3; Ha : < 3 (claim)
25. ⱕ 320 H0: ⱕ 320 (claim); Ha: > 320
26. p ⫽ 0.28 H0 : p ⫽ 0.28 (claim); Ha : p ⫽ 0.28
27. p ⫽ 0.81 H0 : p ⫽ 0.81 (claim); Ha: p ⫽ 0.81
28. < 45 H0 : ⱖ 45; Ha : < 45 (claim)
29. Type I: Rejecting H0: p ⱖ 0.60 when actually p ⱖ 0.60. Type II: Not rejecting H0: p ⱖ 0.60 when actually p < 0.60. 30. Type I: Rejecting H0 : p ⫽ 0.05 when actually p ⫽ 0.05. Type II: Not rejecting H0 : p ⫽ 0.05 when actually p ⫽ 0.05. 31. Type I: Rejecting H0: ⱕ 12 when actually ⱕ 12. Type II: Not rejecting H0: ⱕ 12 when actually > 12. 32. Type I: Rejecting H0 : p ⫽ 0.50 when actually p ⫽ 0.50. Type II: Not rejecting H0 : p ⫽ 0.50 when actually p ⫽ 0.50. 33. Type I: Rejecting H0: p ⫽ 0.88 when actually p ⫽ 0.88. Type II: Not rejecting H0: p ⫽ 0.88 when actually p ⫽ 0.88. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
34. Type I: Rejecting p ⫽ 0.30 when actually p ⫽ 0.30. Type II: Not rejecting H0 : p ⫽ 0.30 when actually p ⫽ 0.30. 35. The null hypothesis is H0: p ⱖ 0.14, the alternative hypothesis is Ha: p < 0.14. Therefore, because the alternative hypothesis contains <, the test is a lefttailed test. 36. The null hypothesis is H0 : ⱕ 0.02, the alternative hypothesis is Ha: > 0.02. Therefore, because the alternative hypothesis contains >, the test is a righttailed test. 37. The null hypothesis is H0: p ⫽ 0.87, the alternative hypothesis is Ha: p ⫽ 0.87. Therefore, because the alternative hypothesis contains ≠, the test is a twotailed test. 38. The null hypothesis is H0 : ⱖ 80,000, the alternative hypothesis is Ha: < 80,000. Therefore, because the alternative hypothesis contains <, the test is a lefttailed test. 39. The null hypothesis is H0: p ⫽ 0.053, the alternative hypothesis is Ha: p ⫽ 0.053. Therefore, because the alternative hypothesis contains ≠, the test is a twotailed test. 40. The null hypothesis is H0 : ⱕ 10, the alternative hypothesis is Ha: > 10. Therefore, because the alternative hypothesis contains >, the test is a righttailed test. 41. (a) There is enough evidence to support the company’s claim. (b) There is not enough evidence to support the company’s claim. 42. (a) There is enough evidence to reject the government worker’s claim. (b) There is not enough evidence to reject the government worker’s claim. 43. (a) There is enough evidence to support the Department of Labor’s claim. (b) There is not enough evidence to support the Department of Labor’s claim. 44. (a) There is enough evidence to reject the manufacturer’s claim. (b) There is not enough evidence to reject the manufacturer’s claim. 45. (a) There is enough evidence to support the manufacturer’s claim. (b) There is not enough evidence to support the manufacturer’s claim. 46. (a) There is enough evidence to reject the softdrink maker’s claim. (b) There is not enough evidence to reject the softdrink maker’s claim. 47. H0 : ⱖ 60; Ha: < 60
48. H0: ⫽ 21; Ha: ⫽ 21
49. (a) H0: ⱖ 15; Ha: < 15
50. (a) H0 : ⱕ 28; Ha : > 28
(b) H0: ⱕ 15 Ha: > 15
(b) H0 : ⱖ 28; Ha : < 28
51. If you decrease ␣, you are decreasing the probability that you reject H0. Therefore, you are increasing the probability of failing to reject H0. This could increase , the probability of failing to reject H0 when H0 is false. 52. If ␣ ⫽ 0, the null hypothesis cannot be rejected and the hypothesis test is useless. 53. (a) Fail to reject H0 because the CI includes values greater than 70. (b) Reject H0 because the CI is located below 70. (c) Fail to reject H0 because the CI includes values greater than 70.
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181
182

CHAPTER 7
HYPOTHESIS TESTING WITH ONE SAMPLE
54. (a) Fail to reject H0 because the CI includes values less than 54. (b) Fail to reject H0 because the CI includes values less than 54. (c) Reject H0 because the CI is located to the right of 54. 55. (a) Reject H0 because the CI is located to the right of 0.20. (b) Fail to reject H0 because the CI includes values less than 0.20. (c) Fail to reject H0 because the CI includes values less than 0.20. 56. (a) Fail to reject H0 because the CI includes values greater than 0.73. (b) Reject H0 because the CI is located to the left of 0.73. (c) Fail to reject H0 because the CI includes values greater than 0.73.
7.2 HYPOTHESIS TESTING FOR THE MEAN (LARGE SAMPLES)
7.2 Try It Yourself Solutions 1a. (1) P ⫽ 0.0347 > 0.01 ⫽ ␣ (2) P ⫽ 0.0347 < 0.05 ⫽ ␣ b. (1) Fail to reject H0 because 0.0347 > 0.01. (2) Reject H0 because 0.0347 < 0.05. 2a.
Area = 0.0526
−3 −2 −1
z 0
z = − 1.62
1
2
3
b. P ⫽ 0.0526 c. Fail to reject H0 because P ⫽ 0.0526 > 0.05 ⫽ ␣. 3a. Area that corresponds to z ⫽ 2.31 is 0.9896.
Area = 0.0104
−3 −2 −1
z 0
1
2
3
z = 2.31
b. P ⫽ 2 共area兲 ⫽ 2共0.0104兲 ⫽ 0.0208 c. Fail to reject H0 because P ⫽ 0.0208 > 0.01 ⫽ ␣.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
4a. The claim is “the mean speed is greater than 35 miles per hour.” H0 : ⱕ 35; Ha: > 35 (claim) b. ␣ ⫽ 0.05 x ⫺ 36 ⫺ 35 1 ⫽ ⫽ ⫽ 2.500 s 4 0.4 冪n 冪100 d. Pvalue ⫽ Area right of z ⫽ 2.50 ⫽ 0.0062 c. z ⫽
e. Reject H0 because Pvalue ⫽ 0.0062 < 0.05 ⫽ ␣. f. Because you reject H0 , there is enough evidence to claim the average speed limit is greater than 35 miles per hour. 5a. The claim is “one of your distributors reports an average of 150 sales per day.” H0 : ⫽ 150 (claim); Ha : ⫽ 150 b. ␣ ⫽ 0.01 x ⫺ 143 ⫺ 150 ⫺7 ⫽ ⫽ ⬇ ⫺2.76 15 2.535 冪n 冪35 d. Pvalue ⫽ 0.0058 c. z ⫽
e. Reject H0 because Pvalue ⫽ 0.0058 < 0.01 ⫽ ␣. f. There is enough evidence to reject the claim. 6a. P ⫽ 0.0440 > 0.01 ⫽ ␣
b. Fail to reject H0.
7a.
8a. 1 α 2
α = 0.10 −3 −2 z 0
z 0
1
2
−3
3
1 α 2
= 0.04 −z 0
0
1 z0
= 0.04 z 3
b. Area ⫽ 0.1003
b. 0.0401 and 0.9599
c. z0 ⫽ ⫺1.28
c. z0 ⫽ ⫺1.75 and 1.75
d. z < ⫺1.28
d. z < ⫺1.75, z > 1.75
9a. The claim is “the mean work day of the firm’s accountants is less than 8.5 hours.” H0: > 8.5; Ha: < 8.5 (claim) b. ␣ ⫽ 0.01 c. z0 ⫽ ⫺2.33; Rejection region: z < ⫺2.33 d. z ⫽
x ⫺ 8.2 ⫺ 8.5 ⫺0.300 ⫽ ⫽ ⬇ ⫺3.550 s 0.5 0.0845 冪n 冪35 Reject H0.
e. α = 0.01 −3
z0
−1
z = − 3.55
z 0
1
2
3
f. There is enough evidence to support the claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
183
184

CHAPTER 7
HYPOTHESIS TESTING WITH ONE SAMPLE
10a. ␣ ⫽ 0.01 b. ± z0 ⫽ ± 2.575; Rejection regions: z < ⫺2.575, z > 2.575 c. Fail to reject H0. 1 α 2
1 α 2
= 0.005
= 0.005
z = −2.24 − z0
−1
0
1
2
z
z0
d. There is not enough evidence to support the claim that the mean cost is significantly different from $10,460 at the 1% level of significance.
7.2 EXERCISE SOLUTIONS 1.
2.
Area = 0.1151
−3 −2 −1
Area = 0.0455 z 0
z = −1.20
1
2
−3 −2 −1
3
z 0
z = −1.69
P ⫽ 0.1151; Fail to reject H0 because P ⫽ 0.1151 > 0.10 ⫽ ␣.
1
2
3
P ⫽ 0.0455; Reject H0 because P ⫽ 0.0455 < 0.05 ⫽ ␣.
3.
4.
Area = 0.0096
−3 −2 −1
Area = 0.1093 z
0
1
2
−3 −2 −1
3
z = 2.34
z 0
1
2
3
z = 1.23
P ⫽ 0.0096; Reject H0 because P ⫽ 0.0096 < 0.01 ⫽ ␣.
P ⫽ 0.1093; Fail to reject H0 because P ⫽ 0.1093 > 0.10 ⫽ ␣.
5.
6.
Area = 0.0594
−3 −2 −1
Area = 0.0107 z 0
z = − 1.56
1
2
−3 −2 −1
3
8. d
1
2
3
z = 2.30
P ⫽ 2共Area兲 ⫽ 2共0.0594兲 ⫽ 0.1188; Fail to reject H0 because P ⫽ 0.1188 > 0.05 ⫽ ␣. 7. c
z 0
9. e
10. f
11. b
P ⫽ 2共Area兲 ⫽ 2共0.0107兲 ⫽ 0.0214; Fail to reject H0 because P ⫽ 0.0214 > 0.01 ⫽ ␣. 12. a
13. (a) Fail to reject H0 . (b) Reject H0 共P ⫽ 0.0461 < 0.05 ⫽ ␣兲.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
14. (a) Fail to reject H0 共P ⫽ 0.0691 > 0.01 ⫽ ␣兲. (b) Fail to reject H0 共P ⫽ 0.0691 > 0.05 ⫽ ␣兲. 15. 1.645
16. 1.41
17. ⫺1.88
18. ⫺1.34
19. ± 2.33
20. ± 1.645
21. Righttailed 共␣ ⫽ 0.01兲
22. Twotailed 共␣ ⫽ 0.05兲
23. Twotailed 共␣ ⫽ 0.10兲
24. Lefttailed 共␣ ⫽ 0.05兲
25. (a) Fail to reject H0 because ⫺1.645 < z < 1.645. (b) Reject H0 because z > 1.645. (c) Fail to reject H0 because ⫺1.645 < z < 1.645. (d) Reject H0 because z < ⫺1.645. 26. (a) Reject H0 because z > 1.96. (b) Fail to reject H0 because ⫺1.96 < z < 1.96. (c) Fail to reject H0 because ⫺1.96 < z < 1.96. (d) Reject H0 because z < ⫺1.96. 27. (a) Fail to reject H0 because z < 1.285. (b) Fail to reject H0 because z < 1.285. (c) Fail to reject H0 because z < 1.285. (d) Reject H0 because z > 1.285. 28. (a) Fail to reject H0 because ⫺2.575 < z < 2.575. (b) Reject H0 because z < ⫺2.575. (c) Reject H0 because z > 2.575. (d) Fail to reject H0 because ⫺2.575 < z < 2.575. 29. H0 : ⫽ 40; Ha : ⫽ 40
⫽ 0.05 → z0 ⫽ ± 1.96 z⫽
x ⫺ 39.2 ⫺ 40 ⫺0.8 ⬇ ⫺2.145 ⫽ ⫽ s 3.23 0.373 冪n 冪75
Reject H0. There is enough evidence to reject the claim. 30. H0 : ⱕ 1030 and Ha : > 1030
␣ ⫽ 0.05 → z0 ⫽ 1.645 z⫽
x ⫺ 1035 ⫺ 1030 5 ⫽ ⫽ ⬇ 1.537 s 23 3.253 冪n 冪50
Fail to reject H0. There is not enough evidence to support the claim.
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31. H0 : ⫽ 6000; Ha : ⫽ 6000
␣ ⫽ 0.01 → z0 ⫽ ± 2.575 z⫽
x ⫺ 5800 ⫺ 6000 ⫺200 ⫽ ⫽ ⬇ ⫺3.381 s 350 59.161 冪n 冪35
Reject H0. There is enough evidence to support the claim. 32. H0 : ⱕ 22,500 and Ha : > 22,500
␣ ⫽ 0.01 → z0 ⫽ 2.33 z⫽
x ⫺ 23,250 ⫺ 22,500 750 ⫽ ⫽ ⬇ 4.193 s 1200 178.885 冪n 冪45
Reject H0 . There is enough evidence to reject the claim. 33. (a) H0 : ⱕ 275; Ha : > 275 (claim) (b) z ⫽
x ⫺ 282 ⫺ 275 7 ⫽ ⫽ ⬇ 1.84 s 35 3796 冪n 冪85
Area ⫽ 0.9671
(c) Pvalue ⫽ 再Area to right of z ⫽ 1.84冎 ⫽ 0.0329 (d) Reject H0 . There is sufficient evidence at the 4% level of significance to support the claim that the mean score for Illinois’ eighth grades is more than 275. 34. (a) H0 : ⱖ 135 (claim); Ha : < 135 (b) z ⫽
x ⫺ 133 ⫺ 135 ⫺2 ⫽ ⫽ ⬇ ⫺3.43 s 3.3 0.583 冪n 冪32
Area ⫽ 0.0003 (c) Pvalue ⫽ 再Area to left of z ⫽ ⫺3.43冎 ⫽ 0.0003 (d) Reject H0 . (e) There is sufficient evidence at the 10% level of significance to reject the claim that the average activating temperature is at least 135⬚F. 35. (a) H0 : ⱕ 8; Ha : > 8 (claim) (b) z ⫽
x ⫺ 7.9 ⫺ 8 ⫺0.1 ⫽ ⫽ ⬇ ⫺0.37 s 2.67 0.267 冪n 冪100
Area ⫽ 0.3557
(c) Pvalue ⫽ 再Area to right of z ⫽ ⫺0.37冎 ⫽ 0.6443 (d) Fail to reject H0 . (e) There is insufficient evidence at the 7% level of significance to support the claim that the mean consumption of tea by a person in the United States is more than 8 gallons per year.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
36. (a) H0 : ⫽ 3.1 (claim); Ha : ⫽ 3.1 (b) z ⫽
x ⫺ 2.9 ⫺ 3.1 ⫺0.2 ⫽ ⫽ ⬇ ⫺1.65 s 0.94 0.121 冪n 冪60
Area ⫽ 0.0495
(c) Pvalue ⫽ 2再Area to left of z ⫽ ⫺1.65冎 ⫽ 2再0.0495冎 ⫽ 0.099 (d) Fail to reject H0. (e) There is insufficient evidence at the 8% level to reject the claim that the mean tuna consumed by a person in the United States is 3.1 pounds per year. 37. (a) H0: ⫽ 15 (claim); Ha : ⫽ 15 (b) x ⬇ 14.834 z⫽
s ⬇ 4.288
x ⫺ 14.834 ⫺ 15 ⫺0.166 ⫽ ⫽ ⬇ ⫺0.219 s 4.288 0.758 冪n 冪32
Area ⫽ 0.4129
(c) Pvalue ⫽ 2再Area to left of z ⫽ ⫺0.22冎 ⫽ 2再0.4129冎 ⫽ 0.8258 (d) Fail to reject H0 . (e) There is insufficient evidence at the 5% level of significance to reject the claim that the mean time it takes smokers to quit smoking permanently is 15 years. 38. (a) H0 : ⱕ $100,800; Ha : > $100,800 (claim) (b) x ⬇ 94,891.47, z⫽
s ⬇ 5239.516
x ⫺ 94,891.47 ⫺ 100,800 ⫺5908.53 ⫽ ⫽ ⬇ ⫺6.58 s 5239.516 898.570 冪n 冪34
Area ⬇ 0
(c) Pvalue ⫽ 再Area to right of z ⫽ ⫺6.58冎 ⬇ 1 (d) Fail to reject H0 . (e) There is insufficient evidence at the 3% level to support the claim that the mean annual salary for engineering managers in Alabama is at least $100,800. 39. (a) H0 : ⫽ 40 (claim); Ha : ⫽ 40 (b) z0 ⫽ ± 2.575; Rejection regions: z < ⫺2.575 and z > 2.575 (c) z ⫽
x ⫺ 39.2 ⫺ 40 ⫺0.8 ⫽ ⫽ ⬇ ⫺0.584 s 7.5 1.369 冪n 冪30
(d) Fail to reject H0. (e) There is insufficient evidence at the 1% level of significance to reject the claim that the mean caffeine content per one 12ounce bottle of cola is 40 milligrams.
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40. (a) H0 : ⫽ 140 (claim); Ha : ⫽ 140 (b) z0 ⫽ ± 1.96; Rejection regions: z < ⫺1.96 and z > 1.96 (c) z ⫽
x ⫺ 146 ⫺ 140 6 ⫽ ⫽ ⬇ 1.77 s 22 3.395 冪n 冪42
(d) Fail to reject H0 . (e) There is insufficient evidence at the 5% level to reject the claim that the mean caffeine content is 140 milligrams per 8 ounces. 41. (a) H0 : ⱖ 750 (claim); Ha : < 750 (b) z0 ⫽ ⫺2.05; Rejection region: z < ⫺2.05 (c) z ⫽
x ⫺ 745 ⫺ 750 ⫺5 ⫽ ⫽ ⬇ ⫺0.500 s 60 10 冪n 冪36
(d) Fail to reject H0 . (e) There is insufficient evidence at the 2% level of significance to reject the claim that the mean life of the bulb is at least 750 hours. 42. (a) H0 : ⱕ 230; Ha : > 230 (claim) (b) z0 ⫽ 1.75; Rejection region: z > 1.75 (c) z ⫽
x ⫺ 232 ⫺ 230 2 ⫽ ⫽ ⬇ 1.44 s 10 1.387 冪n 冪52
(d) Fail to reject H0 . (e) There is insufficient evidence at the 4% level to support the claim that the mean sodium content per serving of cereal is greater than 230 milligrams. 43. (a) H0 : ⱕ 32; Ha : > 32 (claim) (b) z0 ⫽ 1.55; Rejection region: z > 1.55 (c) x ⬇ 29.676 z⫽
s ⬇ 9.164
x ⫺ 29.676 ⫺ 32 ⫺2.324 ⫽ ⫽ ⬇ ⫺1.478 s 9.164 1.572 冪n 冪34
(d) Fail to reject H0 . (e) There is insufficient evidence at the 6% level of significance to support the claim that the mean nitrogen dioxide level in Calgary is greater than 32 parts per billion.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
44. (a) H0: ⱖ 10,000 (claim); Ha : < 10,000 (b) z0 ⫽ ⫺1.34; Rejection region: z < ⫺1.34 (c) x ⫽ 9580.9, z⫽
s ⫽ 1722.4
x ⫺ 9580.9 ⫺ 10,000 ⫺419.1 ⫽ ⫽ ⫽ ⫺1.38 s 1722.4 304.5 冪n 冪32
(d) Reject H0. (e) There is sufficient evidence at the 9% level to reject the claim that the mean life of fluorescent lamps is at least 10,000 hours. 45. (a) H0: ⱖ 10 (claim); Ha : < 10 (b) z0 ⫽ ⫺1.88; Rejection region: z < ⫺1.88 (c) x ⬇ 9.780, s ⬇ 2.362 x⫽
x ⫺ 9.780 ⫺ 10 ⫺0.22 ⫽ ⫽ ⬇ ⫺0.51 s 2.362 0.431 冪n 冪30
(d) Fail to reject H0 . (e) There is insufficient evidence at the 3% level to reject the claim that the mean weight loss after 1 month is at least 10 pounds. 46. (a) H0: ⱖ 60; Ha : < 60 (claim) (b) z0 ⫽ ⫺2.33; Rejection region: z < ⫺2.33 (c) x ⫽ 49, z⫽
s ⫽ 21.51
x ⫺ 49 ⫺ 60 ⫺11 ⫽ ⫽ ⫽ ⫺3.62 s 21.51 3.042 冪n 冪50
(d) Fail to reject H0 . (e) There is sufficient evidence at the 1% level to support the claim that the mean time it takes an employee to evacuate a building during a fire drill is less than 60 seconds. 47. z ⫽
x ⫺ 11,400 ⫺ 11,500 ⫺100 ⬇ ⫺1.71 ⫽ ⫽ s 320 58.424 冪n 冪30
Pvalue ⫽ 再Area left of z ⫽ ⫺1.71冎 ⫽ 0.0436 Fail to reject H0 because the standardized test statistic z ⫽ ⫺1.71 is greater than the critical value z0 ⫽ ⫺2.33. 48. z ⫽
x ⫺ 22,200 ⫺ 22,000 200 ⫽ ⫽ ⫽ 1.548 s 775 129.2 冪n 冪36
Pvalue ⫽ 再Area right of z ⫽ 1.55冎 ⫽ 0.0606 Fail to reject H0 because Pvalue ⫽ 0.0606 > 0.05 ⫽ ␣.
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49. (a) ␣ ⫽ 0.02; Fail to reject H0. (b) ␣ ⫽ 0.05; Reject H0. (c) z ⫽
x ⫺ 11,400 ⫺ 11,500 ⫺100 ⫽ ⫽ ⬇ ⫺2.21 s 320 45.254 冪n 冪50
Pvalue ⫽ 再Area left of z ⫽ ⫺2.21冎 ⫽ 0.0136 → Fail to reject H0. (d) z ⫽
x ⫺ 11,400 ⫺ 11,500 ⫺100 ⫽ ⫽ ⬇ ⫺3.13 s 320 32 冪n 冪100
Pvalue ⫽ 再Area left of z ⫽ ⫺3.13冎 < 0.0009 → Reject H0. 50. (a) ␣ ⫽ 0.06; Fail to reject H0. (b) ␣ ⫽ 0.07; Reject H0. (c) z ⫽
x ⫺ 22,200 ⫺ 20,000 200 ⫽ ⫽ ⫽ 1.63 s 775 122.5 冪n 冪40
Pvalue ⫽ 再Area right of z ⫽ 1.63冎 ⫽ 0.0516; Fail to reject H0. (d) z ⫽
x ⫺ 22,200 ⫺ 20,000 200 ⫽ ⫽ ⬇ 2.31 s 775 86.6 冪n 冪80
Pvalue ⫽ 再Area right of z ⫽ 2.31冎 ⫽ 0.0104; Reject H0. 51. Using the classical ztest, the test statistic is compared to critical values. The ztest using a Pvalue compares the Pvalue to the level of significance ␣.
7.3 HYPOTHESIS TESTING FOR THE MEAN (SMALL SAMPLES)
7.3 Try It Yourself Solutions 1a. 2.650
b. t0 ⫽ ⫺2.650
2a. 1.860
b. t0 ⫽ ⫹1.860
3a. 2.947
b. t0 ⫽ ± 2.947
4a. The claim is “the mean cost of insuring a 2005 Honda Pilot LX is at least $1350.” H0 : ⱖ $1350 (claim); Ha : < $1350 b. ␣ ⫽ 0.01 and d.f. ⫽ n ⫺ 1 ⫽ 8 c. t0 ⫽ ⫺2.896; Reject H0 if t ⱕ ⫺2.896. d. t ⫽
x ⫺ 1290 ⫺ 1350 ⫺60 ⫽ ⫽ ⬇ ⫺2.571 s 70 23.333 冪n 冪9
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
e. Fail to reject H0 . α = 0.01 t0
t = − 2.571 −1
0
t
1
2
3
f. There is not enough evidence to reject the claim. 5a. The claim is “the mean conductivity of the river is 1890 milligrams per liter.” H0 : ⫽ 1890 (claim); Ha : ⫽ 1890 b. ␣ ⫽ 0.01 and d.f. ⫽ n ⫺ 1 ⫽ 18 c. t0 ⫽ ± 2.878; Reject H0 if t < ⫺2.878 or t > 2.878. d. t ⫽
x ⫺ 2500 ⫺ 1890 610 ⫽ ⫽ ⬇ 3.798 s 700 160.591 冪n 冪19
e. Reject H0. 1 α 2
= 0.005
− 4 − t0
1 α 2
= 0.005 t = 3.798
− 1 0 1 2 t0 4
t
f. There is enough evidence to reject the company’s claim. 6a. t ⫽
x ⫺ 172 ⫺ 185 ⫺13 ⫽ ⫽ ⬇ ⫺2.123 s 15 6.124 冪n 冪6
Pvalue ⫽ 再Area left of t ⫽ ⫺2.123冎 ⬇ 0.0436 b. Pvalue ⫽ 0.0436 < 0.05 ⫽ ␣ c. Reject H0 . d. There is enough evidence to reject the claim.
7.3 EXERCISE SOLUTIONS 1. Identify the level of significance ␣ and the degrees of freedom, d.f. ⫽ n ⫺ 1. Find the critical value(s) using the tdistribution table in the row with n ⫺ 1 d.f. If the hypothesis test is: (1) Lefttailed, use “One Tail, ␣” column with a negative sign. (2) Righttailed, use “One Tail, ␣” column with a positive sign. (3) Twotailed, use “Two Tail, ␣” column with a negative and a positive sign. 2. Identify the claim. State H0 and Ha . Specify the level of significance. Identify the degrees of freedom and sketch the sampling distribution. Determine the critical value(s) and rejection region(s). Find the standardized test statistic. Make a decision and interpret it in the context of the original claim. The population must be normal or nearly normal. 3. t0 ⫽ 1.717
4. t0 ⫽ 2.764
5. t0 ⫽ ⫺2.101
7. t0 ⫽ ± 2.779
8. t0 ⫽ ± 2.262
9. 1.328
11. ⫺2.473
12. ⫺3.106
13. ± 3.747
6. t0 ⫽ ⫺1.771 10. 1.895 14. ± 1.721
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15. (a) Fail to reject H0 because t > ⫺2.086. (b) Fail to reject H0 because t > ⫺2.086. (c) Fail to reject H0 because t > ⫺2.086. (d) Reject H0 because t < ⫺2.086. 16. (a) Fail to reject H0 because ⫺1.372 < t < 1.372. (b) Reject H0 because t < ⫺1.372. (c) Reject H0 because t > 1.372. (d) Fail to reject H0 because ⫺1.372 < t < 1.372. 17. (a) Fail to reject H0 because ⫺2.602 < t < 2.602. (b) Fail to reject H0 because ⫺2.602 < t < 2.602. (c) Reject H0 because t > 2.602. (d) Reject H0 because t < ⫺2.602. 18. (a) Fail to reject H0 because ⫺1.725 < t < 1.725. (b) Reject H0 because t < ⫺1.725. (c) Fail to reject H0 because ⫺1.725 < t < 1.725. (d) Reject H0 because t > 1.725. 19. H0 : ⫽ 15 (claim); Ha : ⫽ 15
␣ ⫽ 0.01 and d.f. ⫽ n ⫺ 1 ⫽ 5 t0 ⫽ ± 4.032 t⫽
x ⫺ 13.9 ⫺ 15 ⫺1.1 ⫽ ⫽ ⬇ ⫺0.834 s 3.23 1.319 冪n 冪6
Fail to reject H0. There is not enough evidence to reject the claim. 20. H0 : ⱕ 25; Ha : > 25 (claim)
␣ ⫽ 0.05 and d.f. ⫽ n ⫺ 1 ⫽ 16 t0 ⫽ 1.746 t⫽
x ⫺ 26.2 ⫺ 25 1.2 ⬇ 2.133 ⫽ ⫽ s 2.32 0.563 冪n 冪17
Reject H0. There is enough evidence to support the claim.
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HYPOTHESIS TESTING WITH ONE SAMPLE
21. H0: ⱖ 8000 (claim); Ha: < 8000
␣ ⫽ 0.01 and d.f. ⫽ n ⫺ 1 ⫽ 24 t0 ⫽ ⫺2.492 t⫽
x ⫺ 7700 ⫺ 8000 ⫺300 ⫽ ⫽ ⬇ ⫺3.333 s 450 90 冪n 冪25
Reject H0. There is enough evidence to reject the claim. 22. H0 : ⫽ 52,200; Ha : ⫽ 52,200 (claim)
␣ ⫽ 0.05 and d.f. ⫽ n ⫺ 1 ⫽ 3 t0 ⫽ ± 3.182 x ⫺ 53,220 ⫺ 52,200 1020 ⫽ ⫽ ⫽ 1.7 s 1200 600 冪n 冪4 Fail to reject H0 . There is not enough evidence to support the claim. t⫽
23. (a) H0: > 100 ; Ha : < 100 (claim) (b) t0 ⫽ ⫺3.747; Reject H0 if t < ⫺3.747. (c) t ⫽
x ⫺ 75 ⫺ 100 ⫺25 ⫽ ⫽ ⬇ ⫺4.472 s 12.50 5.590 冪n 冪5
(d) Reject H0. (e) There is sufficient evidence at the 1% significance level to support the claim that the mean repair cost for damaged microwave ovens is less than $100. 24. (a) H0 : ⱕ $95; Ha : > $95 (claim) (b) t0 ⫽ 3.143; Reject H0 if t > 3.143. (c) t ⫽
x ⫺ 100 ⫺ 95 5 ⫽ ⫽ ⬇ 0.311 s 42.50 16.0635 冪n 冪7
(d) Fail to reject H0 . (e) There is not enough evidence at the 1% significance level to support the claim that the mean repair cost for damaged computers is more than $95. 25. (a) H0: ⱕ 1; Ha : > 1 (claim) (b) t0 ⫽ 1.796; Reject H0 if t > 1.796. (c) t ⫽
x ⫺ 1.46 ⫺ 1 0.46 ⫽ ⫽ ⬇ 5.691 s 0.28 0.081 冪n 冪12
(d) Reject H0. (e) There is sufficient evidence at the 5% significance level to support the claim that the mean waste recycled by adults in the United States is more than 1 pound per person per day. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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26. (a) H0 : ⱕ 4; Ha : > 4 (claim) (b) t0 ⫽ 1.833; Reject H0 if t > 1.833. (c) t ⫽
x ⫺ 4.54 ⫺ 4 0.54 ⫽ ⫽ ⫽ 1.411 s 1.21 0.383 冪n 冪10
(d) Fail to reject H0 . (e) There is not enough evidence at the 5% significance level to support the claim that the mean waste generated by adults in the U.S. is more than 4 pounds per day. 27. (a) H0: ⫽ $25,000 (claim); Ha: ⫽ $25,000 (b) t0 ⫽ ± 2.262; Reject H0 if t < ⫺2.262 or t > 2.262. (c) x ⬇ 25,852.2 t⫽
s ⬇ $3197.1
x ⫺ 25,852.2 ⫺ 25,000 ⫺852.2 ⫽ ⫽ ⬇ 0.843 s 3197.1 1011.0 冪n 冪10
(d) Fail to reject H0. (e) There is insufficient evidence at the 5% significance level to reject the claim that the mean salary for fulltime male workers over age 25 without a high school diploma is $25,000. 28. (a) H0 : ⫽ $19,100 (claim); Ha : ⫽ $19,100 (b) t0 ⫽ ± 2.201; Reject H0 if t < ⫺2.201 or t > 2.201. (c) x ⬇ $18,886.5, t⫽
s ⬇ $1397.4
x ⫺ 18,886.5 ⫺ 19,100 ⫺213.5 ⫽ ⫽ ⬇ ⫺0.529 s 1397.4 403.4 冪n 冪12
(d) Fail to reject H0 . (e) There is not enough evidence at the 5% significance level to reject the claim that the mean annual pay for fulltime female workers over age 25 without high school diplomas is $19,100. 29. (a) H0: ⱖ 3.0; Ha: < 3.0 (claim) (b) x ⫽ 1.925 t⫽
x ⫽ 0.654
x ⫺ 1.925 ⫺ 3.0 ⫺1.075 ⫽ ⫽ ⬇ ⫺7.351 s 0.654 0.146 冪n 冪20
Pvalue ⫽ 再Area left of t ⫽ ⫺7.351冎 ⬇ 0 (c) Reject H0. (e) There is sufficient evidence at the 5% significance level to support the claim that teenage males drink fewer than three 12ounce servings of soda per day.
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HYPOTHESIS TESTING WITH ONE SAMPLE
30. (a) H0 : ⱕ $550; Ha : > $550 (claim) (b) x ⫽ 605, t⫽
s ⫽ 150.8
x ⫺ 605 ⫺ 550 55 ⫽ ⫽ ⬇ 1.787 s 150.8 30.782 冪n 冪24
Pvalue ⫽ 再Area right of t ⫽ 1.787冎 ⫽ 0.0436 (c) Reject H0. (d) There is sufficient evidence at the 5% significance level to support the claim that teachers spend a mean of more than $550 of their own money on school supplies in a year. 31. (a) H0 : ⱖ 32; Ha : < 32 (claim) (b) x ⫽ 30.167 t⫽
s ⫽ 4.004
x ⫺ 30.167 ⫺ 32 ⫺1.833 ⫽ ⫽ ⬇ ⫺1.942 s 4.004 0.944 冪n 冪18
Pvalue ⫽ 再Area left of t ⫽ ⫺1.942冎 ⬇ 0.0344 (c) Fail to reject H0. (e) There is insufficient evidence at the 1% significance level to support the claim that the mean class size for fulltime faculty is fewer than 32. 32. (a) H0 : ⫽ 11.0 (claim); Ha : ⫽ 11.0 (b) x ⫽ 10.050, s ⫽ 2.485 t⫽
x ⫺ 10.050 ⫺ 11.0 ⫺.95 ⫽ ⫽ ⬇ ⫺1.081 s 2.485 0.879 冪n 冪8
Pvalue ⫽ 2再area left of t ⫽ ⫺1.081冎 ⫽ 2共0.15775兲 ⫽ 0.3155 (c) Fail to reject H0 . (d) There is not enough evidence at the 1% significance level to reject the claim that the mean number of classroom hours per week for fulltime faculty is 11.0. 33. (a) H0 : ⫽ $2634 (claim); Ha : ⫽ $2634 (b) x ⫽ $2785.6 t⫽
s ⫽ $759.3
x ⫺ 2785.6 ⫺ 2634 151.6 ⫽ ⫽ ⫽ 0.692 s 759.3 219.19 冪n 冪12
Pvalue ⫽ 2再Area right of t ⫽ 0.692冎 ⫽ 2再0.2518冎 ⫽ 0.5036 (c) Fail to reject H0. (e) There is insufficient evidence at the 2% significance level to reject the claim that the typical household in the U.S. spends a mean amount of $2634 per year on food away from home.
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34. (a) H0 : ⫽ $152 (claim); Ha : ⫽ $152 (b) x ⫽ $142.8, s ⫽ $37.52 t⫽
x ⫺ 142.8 ⫺ 152 ⫺9.2 ⫽ ⫽ ⫽ ⫺0.775 s 37.52 11.865 冪n 冪10
Pvalue ⫽ 2再Area left of t ⫽ ⫺0.78冎 ⫽ 2再0.229冎 ⫽ 0.4580 (c) Fail to reject H0. (d) There is insufficient evidence at the 2% significance level to reject the claim that the daily lodging costs for a family in the U.S. is $152. 35. H0 : ⱕ $2328; Ha : > 2328 (claim) t⫽
x ⫺ 2528 ⫺ 2328 200 ⫽ ⫽ ⬇ 1.507 s 325 132.681 冪n 冪6
Pvalue ⫽ 再Area right of t ⫽ 1.507冎 ⬇ 0.096 Because 0.096 > 0.01 ⫽ ␣, fail to reject H0 . 36. (a) Because 0.096 > 0.05 ⫽ ␣, fail to reject H0. (b) Because 0.096 < 0.10 ⫽ ␣, reject H0. (c) t ⫽
x ⫺ 2528 ⫺ 2328 200 ⫽ ⫽ ⫽ 2.132 s 325 93.819 冪n 冪12
Pvalue ⫽ 再Area right of t ⫽ 2.132冎 ⬇ 0.028 Because 0.028 > 0.01 ⫽ ␣, fail to reject H0. (d) t ⫽
200 x ⫺ 2528 ⫺ 2328 ⫽ ⫽ ⫽ 3.015 s 325 66.340 冪n 冪24
Pvalue ⫽ 再Area right of t ⫽ 3.015冎 ⬇ 0.003 Since 0.003 < 0.01 ⫽ ␣, reject H0. 37. Because is unknown, n < 30, and the gas mileage is normally distributed, use the tdistribution. H0 : ⱖ 23 (claim); Ha : < 23 t⫽
x ⫺ 22 ⫺ 23 ⫺1 ⫽ ⫽ ⬇ ⫺0.559 s 4 1.789 冪n 冪5
Pvalue ⫽ 再Area left of t ⫽ ⫺0.559冎 ⫽ 0.303 Fail to reject H0. There is insufficient evidence at the 5% significance level to reject the claim that the mean gas mileage for the luxury sedan is at least 23 miles per gallon.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
38. Because is unknown and n ⱖ 30, use the zdistribution. H0 : ⱖ 23,000; Ha : < 23,000 (claim) z⫽
x ⫺ 21,856 ⫺ 23,000 ⫺1144 ⫽ ⫽ ⬇ ⫺2.557 s 3163 447.32 冪n 冪50
Pvalue ⫽ 2再Area left of z ⫽ ⫺2.557冎 ⫽ 2共0.0026兲 ⫽ 0.0052 Reject H0 . There is enough evidence at the 1% significance level to reject the claim that the mean price for 1 year of graduate school for a fulltime student in a master’s degree program at a public institution is less than $23,000.
7.4 HYPOTHESIS TESTING FOR PROPORTIONS
7.4 Try It Yourself Solutions 1a. np ⫽ 共86兲共0.30兲 ⫽ 25.8 > 5, nq ⫽ 共86兲共0.70兲 ⫽ 60.2 > 5 b. The claim is “less than 30% of cellular phone users whose phone can connect to the Internet have done so while at home.” H0 : p ⱖ 0.30; Ha : p < 0.30 (claim) c. ␣ ⫽ 0.05 d. z0 ⫽ ⫺1.645; Reject H0 if z < ⫺1.645. p⫺p ^
e. z ⫽
⫽
0.20 ⫺ 0.30
冪pqn 冪共0.3086兲共0.70兲
⫽
⫺0.1 ⬇ ⫺2.024 0.0494
f. Reject H0 . g. There is enough evidence to support the claim. 2a. np ⫽ 共250兲共0.05兲 ⫽ 12.5 > 5, nq ⫽ 共250兲共0.95兲 ⫽ 237.5 > 5 b. The claim is “5% of U.S. adults have had vivid dreams about UFOs.” H0 : p ⫽ 0.05 (claim); Ha: p ⫽ 0.05 c. ␣ ⫽ 0.01 d. z0 ⫽ ± 2.575; Reject H0 if z < ⫺2.575 or z > 2.575. p⫺p ^
e. z ⫽
⫽
0.08 ⫺ 0.05
兲共0.95兲 冪pqn 冪共0.05250
⫽
0.03 ⬇ 2.176 0.0138
f. Fail to reject H0 . g. There is not enough evidence to reject the claim. 3a. np ⫽ 共75兲共0.30兲 ⫽ 22.5 > 5, nq ⫽ 共75兲共0.70兲 ⫽ 52.5 > 5 b. The claim is “more than 30% of U.S. adults regularly watch the Weather Channel.” H0 : p ⱕ 0.30; Ha: p > 0.30 (claim) © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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c. ␣ ⫽ 0.01 d. z0 ⫽ 2.33; Reject H0 if z > 2.33. x 27 ⫽ ⫽ 0.360 n 75 p⫺p 0.360 ⫺ 0.30 0.06 z⫽ ⫽ ⫽ ⬇ 1.13 0.053 pq 共0.30兲共0.70兲 n 75
e. p ⫽ ^
^
冪
冪
f. Fail to reject H0 . g. There is not enough evidence to support the claim.
7.4 EXERCISE SOLUTIONS 1. Verify that np ⱖ 5 and nq ⱖ 5. State H0 and Ha . Specify the level of significance ␣. Determine the critical value(s) and rejection region(s). Find the standardized test statistic. Make a decision and interpret in the context of the original claim. 2. If np ⱖ 5 and nq ⱖ 5, the normal distribution can be used. 3. np ⫽ 共105兲共0.25兲 ⫽ 26.25 > 5 nq ⫽ 共105兲共0.75兲 ⫽ 78.75 > 5 → use normal distribution H0 : p ⫽ 0.25; Ha : p ⫽ 0.25 (claim) z0 ± 1.96 p⫺p ^
z⫽
⫽
0.239 ⫺ 0.25
兲共0.75兲 冪pqn 冪共0.25105
⫽
⫺0.011 ⬇ ⫺0.260 0.0423
Fail to reject H0 . There is not enough evidence to support the claim. 4. np ⫽ 共500兲共0.30兲 ⫽ 150 ⱖ 5 nq ⫽ 共500兲共0.70兲 ⫽ 350 ⱖ 5 → use normal distribution H0 : p ⱕ 0.30 (claim); Ha : p > 0.30 z0 ⫽ 1.645 p⫺p ^
z⫽
⫽
0.35 ⫺ 0.30
兲共0.70兲 冪pqn 冪共0.30500
⫽
0.05 ⬇ 2.440 0.0205
Reject H0 . There is enough evidence to reject the claim. 5. np ⫽ 共20兲共0.12兲 ⫽ 2.4 < 5 nq ⫽ 共20兲共0.88兲 ⫽ 17.6 ⱖ 5 → cannot use normal distribution
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
6. np ⫽ 共45兲共0.125兲 ⫽ 5.625 ⱖ 5 nq ⫽ 共45兲共0.875兲 ⫽ 39.375 ⱖ 5 → use normal distribution H0 : p ⱕ 0.125; Ha : p > 0.125 (claim) z0 ⫽ 2.33 p⫺p ^
z⫽
⫽
0.2325 ⫺ 0.125
冪pqn 冪共0.12545兲共0.875兲
⫽
0.1075 ⬇ 2.180 0.0493
Fail to reject H0. There is not enough evidence to support the claim. 7. np ⫽ 共70兲共0.48兲 ⫽ 33.6 ⱖ 5 nq ⫽ 共70兲共0.52兲 ⫽ 36.4 ⱖ 5 → use normal distribution H0 : p ⱖ 0.48 (claim); Ha : p < 0.48 z0 ⫽ ⫺1.29 p⫺p ^
z⫽
⫽
0.40 ⫺ 0.48
冪pzn 冪共0.4870兲共0.52兲
⫽
⫺0.08 ⬇ ⫺1.34 0.060
Reject H0 . There is enough evidence to reject the claim. 8. np ⫽ 共16兲共0.80兲 ⫽ 12.8 ⱖ 5 nq ⫽ 共16兲共0.20兲 ⫽ 3.2 < 5 → cannot use normal distribution 9. (a) H0 : p ⱖ 0.20 (claim); Ha : p < 0.20 (b) z0 ⫽ ⫺2.33; Reject H0 if z < ⫺2.33. p⫺p ^
(c) z ⫽
⫽
0.185 ⫺ 0.20
兲共0.80兲 冪pqn 冪共0.20200
⫽
⫺0.015 ⬇ ⫺0.53 0.0283
(d) Fail to reject H0. (e) There is insufficient evidence at the 1% significance level to reject the claim that at least 20% of U.S. adults are smokers. 10. (a) H0 : p ⱕ 0.40 (claim); Ha : p > 0.40 (b) z0 ⫽ 2.33; Reject H0 if z > 2.33. p⫺p ^
(c) z ⫽
⫽
0.416 ⫺ 0.40
兲共0.60兲 冪pqn 冪共0.40250
⫽
0.016 ⬇ 0.52 0.0310
(d) Fail to reject H0 . (e) There is not enough evidence at the 1% significance level to reject the claim that no more than 40% of U.S. adults eat breakfast every day.
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11. (a) H0 : p ⱕ 0.30; Ha : p > 0.30 (claim) (b) z0 ⫽ 1.88; Reject H0 if z > 1.88. p⫺p
0.32 ⫺ 0.3
^
(c) z ⫽
⫽
兲共0.70兲 冪pqn 冪共0.301050
⫽
0.02 ⬇ 1.41 0.0141
(d) Fail to reject H0. (e) There is insufficient evidence at the 3% significance level to support the claim that more than 30% of U.S. consumers have stopped buying the product because the manufacturing of the product pollutes the environment. 12. (a) H0 : p ⱕ 0.60; Ha : p > 0.60 (claim) (b) z0 ⫽ 1.28; Reject H0 if z > 1.28. p⫺p ^
(c) z ⫽
⫽
0.65 ⫺ 0.60
兲共0.40兲 冪pqn 冪共0.60100
⫽
0.05 ⫽ 1.02 0.0490
(d) Fail to reject H0 . (e) There is not enough evidence at the 10% significance level to support the claim that more than 60% of British consumers are concerned about the use of genetic modification in food production and want to avoid genetically modified foods. 13. (a) H0 : p ⫽ 0.44 (claim); Ha : p ⫽ 0.44 (b) z0 ⫽ ± 2.33; Reject H0 if z < ⫺2.33 or z > 2.33. (c) p ⫽ ^
722 ⬇ 0.410 1762 p⫺p ^
z⫽
⫽
0.410 ⫺ 0.44
兲共0.56兲 冪pqn 冪共0.441762
⫽
⫺0.03024 ⬇ ⫺2.537 0.01183
(d) Reject H0. (e) There is sufficient evidence at the 2% significance level to reject the claim that 44% of home buyers find their real estate agent through a friend. 14. (a) H0 : p ⫽ 0.24 (claim); Ha : p ⫽ 0.24 (b) z0 ⫽ ± 1.96; Reject H0 if z < ⫺1.96 or z > 1.96. (c) p ⫽ ^
292 ⬇ 0.2716 1075 p⫺p ^
z⫽
⫽
0.2716 ⫺ 0.240
兲共0.760兲 冪pqn 冪共0.2401075
⫽
0.316 ⬇ 2.43 0.013
(d) Reject H0 . (e) There is sufficient evidence at the 5% significance level to reject the claim that 24% of adults in the United States are afraid to fly.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
15. H0 : p ⱖ 0.52 (claim); Ha : p < 0.52 z0 ⫽ ⫺1.645; Rejection region: z < ⫺1.645 p⫺p ^
z⫽
⫽
0.48 ⫺ 0.52
冪pqn 冪共0.5250兲共0.48兲
⫽
⫺0.04 ⬇ ⫺0.566 0.0707
Fail to reject H0 . There is insufficient evidence to reject the claim. 16. The company should continue the use of giveaways because there is not enough evidence to say that less than 52% of the adults would be more likely to buy a product when there are free samples. 17. H0 : p ⫽ 0.44 (claim); Ha : p ⫽ 0.44 z⫽
x ⫺ np 722 ⫺ 共1762兲共0.44兲 ⫺53.28 ⫽ ⫽ ⬇ ⫺2.56 20.836 冪npq 冪共1762兲共0.44兲共0.56兲
Reject H0. The results are the same. p⫺p ^
18. z ⫽
冪pqn
⇒
冢nx 冣 ⫺ p
冪pqn
⇒
冢nx 冣 ⫺ p 冪pq
冤 冢nx 冣 ⫺ p冥 ⇒ 冪pq
冪n
x ⫺ p冥 n 冤 冢 n n冣 ⇒
冪
⭈冪
n
冪pqn
⇒
x ⫺ np 冪pqn
冪n
7.5 HYPOTHESIS TESTING FOR VARIANCE AND STANDARD DEVIATION
7.5 Try It Yourself Solutions 1a. 20 ⫽ 33.409 2a. 20 ⫽ 17.708 3a. R2 ⫽ 31.526
b. L2 ⫽ 8.231
4a. The claim is “the variance of the amount of sports drink in a 12ounce bottle is no more than 0.40.” H0 : 2 ⱕ 0.40 (claim); Ha : 2 > 0.40 b. ␣ ⫽ 0.01 and d.f. ⫽ n ⫺ 1 ⫽ 30 c. 20 ⫽ 50.892; Reject H0 if 2 > 50.892.
共n ⫺ 1兲s2 共30兲共0.75兲 ⫽ ⫽ 56.250 2 0.40 e. Reject H0 . d. 2 ⫽
f. There is enough evidence to reject the claim.
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5a. The claim is “the standard deviation in the length of response times is less than 3.7 minutes.” H0 : ⱖ 3.7; Ha : < 3.7 (claim) b. ␣ ⫽ 0.05 and d.f. ⫽ n ⫺ 1 ⫽ 8 c. 20 ⫽ 2.733; Reject H0 if 2 < 2.733. d. 2 ⫽
共n ⫺ 1兲s2 共8兲共3.0兲2 ⫽ ⬇ 5.259 2 共3.7兲2
e. Fail to reject H0 . f. There is not enough evidence to support the claim. 6a. The claim is “the variance of the diameters in a certain tire model is 8.6.” H0 : 2 ⫽ 8.6 (claim); Ha : 2 ⫽ 8.6 b. ␣ ⫽ 0.01 and d.f. ⫽ n ⫺ 1 ⫽ 9 c. 2L ⫽ 1.735 and R2 ⫽ 23.589 Reject H0 if 2 > 23.589 or 2 < 1.735. d. 2 ⫽
共n ⫺ 1兲s 2 共9兲共4.3兲 ⫽ ⫽ 4.50 2 共8.6兲
e. Fail to reject H0 . f. There is not enough evidence to reject the claim.
7.5 EXERCISE SOLUTIONS 1. Specify the level of significance ␣. Determine the degrees of freedom. Determine the critical values using the 2 distribution. If (a) righttailed test, use the value that corresponds to d.f. and ␣ . (b) lefttailed test, use the value that corresponds to d.f. and 1 ⫺ ␣ ; and 1 1 (c) twotailed test, use the value that corresponds to d.f. and 2␣ and 1 ⫺ 2␣. 2. State H0 and Ha . Specify the level of significance. Determine the degrees of freedom. Determine the critical value(s) and rejection region(s). Find the standardized test statistic. Make a decision and interpret in the context of the original claim. 3. 20 ⫽ 38.885
4. 20 ⫽ 14.684
7. L2 ⫽ 7.261, R2 ⫽ 24.996
5. 20 ⫽ 0.872
6. 20 ⫽ 13.091
8. L2 ⫽ 12.461, R2 ⫽ 50.993
9. (a) Fail to reject H0.
10. (a) Fail to reject H0.
(b) Fail to reject H0.
(b) Fail to reject H0.
(c) Fail to reject H0.
(c) Reject H0.
(d) Reject H0.
(d) Reject H0.
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CHAPTER 7
11. (a) Fail to reject H0.

HYPOTHESIS TESTING WITH ONE SAMPLE
12. (a) Fail to reject H0.
(b) Reject H0.
(b) Fail to reject H0.
(c) Reject H0.
(c) Fail to reject H0.
(d) Fail to reject H0.
(d) Reject H0.
13. H0: 2 ⫽ 0.52 (claim); Ha: 2 ⫽ 0.52
L2 ⫽ 7.564, R2 ⫽ 30.191 2 ⫽
共n ⫺ 1兲s2 共17兲共0.508兲2 ⫽ ⬇ 16.608 2 共0.52兲
Fail to reject H0. There is insufficient evidence to reject the claim. 14. H0: ⱖ 40; Ha: < 40 (claim)
20 ⫽ 3.053 2 ⫽
共n ⫺ 1兲s2 共11兲共40.8兲2 ⫽ ⬇ 11.444 2 共40兲2
Fail to reject H0 . There is insufficient evidence to support the claim. 15. (a) H0: 2 ⫽ 3 (claim); Ha: 2 ⫽ 3 (b) L2 ⫽ 13.844, R2 ⫽ 41.923; Reject H0 if 2 > 41.923 or 2 < 13.844. (c) 2 ⫽
共n ⫺ 1兲s2 共26兲共2.8兲 ⫽ ⬇ 24.267 2 3
(d) Fail to reject H0. (e) There is insufficient evidence at the 5% level of significance to reject the claim that the variance of the life of the appliances is 3. 16. (a) H0 : 2 ⫽ 6 (claim); Ha : 2 ⫽ 6 (b) L2 ⫽ 14.573, R2 ⫽ 43.194; Reject H0 if 2 > 43.194 or 2 < 14.573. (c) 2 ⫽
共n ⫺ 1兲s2 共27兲共4.25兲 ⫽ ⫽ 19.125 2 6
(d) Fail to reject H0 . (e) There is not enough evidence at the 5% significance level to reject the claim that the variance of the gas mileage is 6. 17. (a) H0: ⱖ 36; Ha: < 36 (claim) (b) 20 ⫽ 13.240; Reject H0 if 2 < 13.240. (c) 2 ⫽
共n ⫺ 1兲s2 共21兲共33.4兲2 ⫽ ⬇ 18.076 2 共36兲2
(d) Fail to reject H0. (e) There is insufficient evidence at the 10% significance level to support the claim that the standard deviation for eighth graders on the examination is less than 36.
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18. (a) H0 : ⱖ 30; Ha : < 30 (claim) (b) 20 ⫽ 6.408; Reject H0 if 2 < 6.408. (c) 2 ⫽
共n ⫺ 1兲s2 共17兲共33.6兲2 ⫽ ⬇ 21.325 2 共30兲2
(d) Fail to reject H0 . (e) There is not enough evidence at the 1% significance level to support the claim that the standard deviation of test scores for eighth grade students who took a U.S. history assessment test is less than 30 points. 19. (a) H0: ⱕ 0.5 (claim); Ha: > 0.5 (b) 20 ⫽ 33.196; Reject H0 if 2 > 33.196. (c) 2 ⫽
共n ⫺ 1兲s2 共24兲共0.7兲2 ⫽ ⫽ 47.04 2 共0.5兲2
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to reject the claim that the standard deviation of waiting times is no more than 0.5 minute. 20. (a) H0 : ⫽ 6.14 (claim); Ha: ⫽ 6.14 (b) L2 ⫽ 8.907, R2 ⫽ 32.852; Reject H0 if 2 < 8.907 or 2 > 32.852. (c) 2 ⫽
共n ⫺ 1兲s 2 共19兲共6.5兲2 ⫽ ⫽ 21.293 2 共6.14兲2
(d) Fail to reject H0 . (e) There is insufficient evidence at the 5% significance level to reject the claim that the standard deviation of the lengths of stay is 6.14 days. 21. (a) H0: ⱖ $3500; Ha: < $3500 (claim) (b) 20 ⫽ 18.114; Reject H0 if 2 < 18.114. (c) 2 ⫽
共n ⫺ 1兲s2 共27兲共4100兲2 ⫽ ⬇ 37.051 2 共3500兲2
(d) Fail to reject H0. (e) There is insufficient evidence at the 10% significance level to support the claim that the standard deviation of the total charge for patients involved in a crash where the vehicle struck a construction baracade is less than $3500. 22. (a) H0 : ⱕ $30 (claim); Ha : > $30 (b) 20 ⫽ 37.566; Reject H0 if 2 < 37.566. (c) 2 ⫽
共n ⫺ 1兲s2 共20兲共35.25兲2 ⫽ ⫽ 27.613 2 共30兲2
(d) Fail to reject H0 . (e) There is not enough evidence at the 1% significance level to reject the claim that the standard deviation of the room rates of hotels in the city is no more than $30.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
23. (a) H0: ⱕ $20,000; Ha: > $20,000 (claim) (b) 20 ⫽ 24.996; Reject H0 if 2 > 24.996. (c) s ⫽ 20,826.145
2 ⫽
共n ⫺ 1兲s 2 共15兲共20,826.145兲2 ⫽ ⬇ 16.265 2 共20,000兲2
(d) Fail to reject H0 . (e) There is insufficient evidence at the 5% significance level to support the claim that the standard deviation of the annual salaries for actuaries is more than $20,000. 24. (a) H0 : ⱖ $14,500 (claim); Ha : < $14,500 (b) 20 ⫽ 10.085; Reject H0 if 2 < 10.085. (c) s ⫽ 13,950.604
2 ⫽
共n ⫺ 1兲s2 共17兲共13,950.604兲2 ⫽ ⬇ 15.736 2 共14,500兲2
(d) Fail to reject H0 . (e) There is not enough evidence at the 10% significance level to reject the claim that the standard deviation of the annual salaries for public relations managers is at least $14,500. 25. 2 ⫽ 37.051 Pvalue ⫽ 再Area left of 2 ⫽ 37.051冎 ⫽ 0.9059 Fail to reject H0 because Pvalue ⫽ 0.9059 > 0.10 ⫽ ␣. 26. 2 ⫽ 27.613 Pvalue ⫽ 再Area right of 2 ⫽ 27.613冎 ⫽ 0.1189 Fail to reject H0 because Pvalue ⫽ 0.1189 > 0.01 ⫽ ␣. 27. 2 ⫽ 16.265 Pvalue ⫽ 再Area right of 2 ⫽ 16.265冎 ⫽ 0.3647 Fail to reject H0 because Pvalue ⫽ 0.3647 > 0.05 ⫽ ␣. 28. 2 ⫽ 15.736 Pvalue ⫽ 再Area left of 2 ⫽ 15.736冎 ⫽ 0.4574 Fail to reject H0 because Pvalue ⫽ 0.4574 > 0.10 ⫽ ␣.
CHAPTER 7 REVIEW EXERCISE SOLUTIONS 1. H0 : ⱕ 1479 (claim); Ha : > 1479
2. H0 : ⫽ 95 (claim); Ha : ⫽ 95
3. H0 : p ⱖ 0.205; Ha : p < 0.205 (claim)
4. H0 : ⫽ 150,020; Ha : ⫽ 150,020 (claim)
5. H0 : ⱕ 6.2; Ha : > 6.2 (claim)
6. H0 : p ⱖ 0.78 (claim); Ha : p < 0.78
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7. (a) H0 : p ⫽ 0.73 (claim); Ha : p ⫽ 0.73 (b) Type I error will occur if H0 is rejected when the actual proportion of college students that occasionally or frequently come late to class is 0.63. Type II error if H0 is not rejected when the actual proportion of college students that occasionally or frequently come late to class is not 0.63. (c) Twotailed, because hypothesis compares “⫽ vs ⫽”. (d) There is enough evidence to reject the claim. (e) There is not enough evidence to reject the claim. 8. (a) H0 : ⱖ 30,000 (claim); Ha : < 30,000 (b) Type I error will occur if H0 is rejected when the actual mean tire life is at least 30,000 miles. Type II error if H0 is not rejected when the actual mean tire life is less than 30,000 miles. (c) Lefttailed, because hypothesis compares “ ⱖ vs <”. (d) There is enough evidence to reject the claim. (e) There is not enough evidence to reject the claim. 9. (a) H0 : ⱕ 50 (claim); Ha : > 50 (b) Type I error will occur if H0 is rejected when the actual standard deviation sodium content is no more than 50 milligrams. Type II error if H0 is not rejected when the actual standard deviation sodium content is more than 50 milligrams. (c) Righttailed, because hypothesis compares “ ⱕ vs >”. (d) There is enough evidence to reject the claim. (e) There is not enough evidence to reject the claim. 10. (a) H0 : ⱖ 25; Ha : < 25 (claim) (b) Type I error will occur if H0 is rejected when the actual mean number of grams of carbohydrates in one bar is greater than or equal to 25. Type II error if H0 is not rejected when the actual mean number of grams of carbohydrates in one bar is less than 25. (c) Lefttailed, because hypothesis compares “ ⱖ vs <”. (d) There is enough evidence to support the claim. (e) There is not enough evidence to support the claim. 11. z0 ⬇ ⫺2.05
12. z0 ⫽ ± 2.81
13. z0 ⫽ 1.96
14. z0 ⫽ ± 1.75
15. H0 : ⱕ 45 (claim); Ha : > 45 z0 ⫽ 1.645 z⫽
x ⫺ 47.2 ⫺ 45 2.2 ⫽ ⫽ ⬇ 2.128 s 6.7 1.0338 冪n 冪42
Reject H0 . There is enough evidence to reject the claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
16. H0 : ⫽ 0; Ha : ⫽ 0 (claim) z0 ⫽ ± 1.96 z⫽
x ⫺ ⫺0.69 ⫺ 0 ⫺0.69 ⫽ ⫽ ⬇ ⫺2.040 s 2.62 0.338 冪n 冪60
Reject H0. There is enough evidence to support the claim. 17. H0: ⱖ 5.500; Ha: < 5.500 (claim) z0 ⫽ ⫺2.33 z⫽
x ⫺ 5.497 ⫺ 5.500 ⫺0.003 ⫽ ⫽ ⬇ ⫺1.636 s 0.011 0.00183 冪n 冪36
Fail to reject H0. There is not enough evidence to support the claim. 18. H0 : ⫽ 7450 (claim); Ha : ⫽ 7450 z0 ⫽ ± 1.96 z⫽
x ⫺ 7512 ⫺ 7450 62 ⫽ ⫽ ⬇ 1.926 s 243 32.186 冪n 冪57
Fail to reject H0 . There is not enough evidence to reject the claim. 19. H0: ⱕ 0.05 (claim); Ha: > 0.05 z⫽
x ⫺ 0.057 ⫺ 0.05 0.007 ⫽ ⫽ ⬇ 2.20 s 0.018 0.00318 冪n 冪32
Pvalue ⫽ 再Area right of z ⫽ 2.20冎 ⫽ 0.0139
␣ ⫽ 0.10; Reject H0. ␣ ⫽ 0.05; Reject H0. ␣ ⫽ 0.01; Fail to reject H0. 20. H0 : ⫽ 230; Ha : ⫽ 230 (claim) z⫽
x ⫺ 216.5 ⫺ 230 ⫺13.5 ⫽ ⫽ ⬇ ⫺5.41 s 17.3 2.497 冪n 冪48
Pvalue ⫽ 2再Area left of z ⫽ ⫺5.41冎 ⬇ 0
␣ ⫽ 0.10; Reject H0. ␣ ⫽ 0.05; Reject H0. ␣ ⫽ 0.01; Reject H0.
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21. H0: ⫽ 326 (claim); Ha: ⫽ 326 z⫽
x ⫺ 318 ⫺ 326 ⫺8 ⫽ ⫽ ⬇ ⫺2.263 s 25 3.536 冪n 冪50
Pvalue ⫽ 2再Area left of z ⫽ ⫺2.263冎 ⫽ 2再0.012冎 ⫽ 0.024 Reject H0. There is sufficient evidence to reject the claim. 22. H0 : ⱕ $650 (claim); Ha : > $650 z⫽
x ⫺ 657 ⫺ 650 7 ⫽ ⫽ ⬇ 1.174 s 40 5.963 冪n 冪45
Pvalue ⫽ 再Area right of z ⫽ 1.17冎 ⫽ 0.1210 Fail to reject H0 .There is not enough evidence to reject the claim. 23. t0 ⫽ ± 2.093
24. t0 ⫽ 2.998
25. t0 ⫽ ⫺1.345
26. t0 ⫽ ± 2.201
27. H0: ⫽ 95; Ha: ⫽ 95 (claim) t0 ⫽ ± 2.201 t⫽
x ⫺ 94.1 ⫺ 95 ⫺0.9 ⫽ ⫽ ⬇ ⫺2.038 s 1.53 0.442 冪n 冪12
Fail to reject H0. There is not enough evidence to support the claim. 28. H0 : ⱕ 12,700; Ha : > 12,700 (claim) t0 ⫽ 1.725 t⫽
x ⫺ 12,804 ⫺ 12,700 104 ⫽ ⫽ ⬇ 1.922 s 248 54.118 冪n 冪21
Reject H0. There is enough evidence to support the claim. 29. H0: ⱖ 0 (claim); Ha: < 0 t0 ⫽ ⫺1.341 t⫽
x ⫺ ⫺0.45 ⫺ 0 ⫺0.45 ⬇ ⫺1.304 ⫽ ⫽ s 1.38 0.345 冪n 冪16
Fail to reject H0. There is not enough evidence to reject the claim. 30. H0 : ⫽ 4.20 (claim); Ha : ⫽ 4.20 t0 ⫽ ± 2.896 t⫽
x ⫺ 4.41 ⫺ 4.20 0.21 ⫽ ⫽ ⬇ 2.423 s 0.26 0.0867 冪n 冪9
Fail to reject H0 . There is not enough evidence to reject the claim.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
31. H0: ⱕ 48 (claim); Ha: > 48 t0 ⫽ 3.148 t⫽
x ⫺ 52 ⫺ 48 4 ⫽ ⫽ ⬇ 4.233 s 2.5 0.945 冪n 冪7
Reject H0. There is enough evidence to reject the claim. 32. H0 : ⱖ 850; Ha : < 850 (claim) t0 ⫽ ⫺2.160 t⫽
x ⫺ 875 ⫺ 850 25 ⫽ ⫽ ⬇ 3.742 s 25 6.682 冪n 冪14
Fail to reject H0 . There is not enough evidence to support the claim. 33. H0 : ⫽ $25 (claim); Ha : ⫽ $25 t0 ⫽ ± 1.740 t⫽
x ⫺ 26.25 ⫺ 25 1.25 ⫽ ⫽ ⬇ 1.642 s 3.23 0.761 冪n 冪18
Fail to reject H0 . There is not enough evidence to reject the claim. 34. H0 : ⱕ 10 (claim); Ha : > 10 t0 ⫽ 1.397 t⫽
x ⫺ 13.5 ⫺ 10 3.5 ⫽ ⫽ ⬇ 1.810 s 5.8 1.933 冪n 冪9
Reject H0. There is enough evidence to reject the claim. 35. H0 : ⱖ $10,200 (claim); Ha: < $10,200 t0 ⫽ ⫺2.602 x ⫽ 9895.8 t⫽
s ⫽ 490.88
x ⫺ 9895.8 ⫺ 10,200 ⫺304.2 ⫽ ⫽ ⬇ ⫺2.479 s 490.88 122.72 冪n 冪16
Pvalue ⬇ 0.0128 Fail to reject H0 . There is not enough evidence to reject the claim. 36. H0 : ⱕ 9; Ha : > 9 (claim) x ⫽ 9.982, t⫽
s ⫽ 2.125
x ⫺ 9.982 ⫺ 9 0.982 ⫽ ⫽ ⬇ 1.532 s 2.125 0.641 冪n 冪11
Pvalue ⫽ 0.078 Fail to reject H0 . There is not enough evidence to support the claim.
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37. H0 : p ⫽ 0.15 (claim); Ha : p ⫽ 0.15 z0 ⫽ ± 1.96 p⫺p ^
z⫽
⫽
0.09 ⫺ 0.15
冪pqn 冪共0.1540兲共0.85兲
⫽
⫺0.06 ⬇ ⫺1.063 0.0565
Fail to reject H0 . There is not enough evidence to reject the claim. 38. H0 : p ⱖ 0.70; Ha : p < 0.70 (claim) z0 ⫽ ⫺2.33 p⫺p ^
z⫽
⫽
0.50 ⫺ 0.70
冪pqn 冪共0.7068兲共0.30兲
⫽
⫺0.2 ⬇ ⫺3.599 0.0556
Reject H0 . There is enough evidence to support the claim. 39. Because np ⫽ 3.6 is less than 5, the normal distribution cannot be used to approximate the binomial distribution. 40. H0 : p ⫽ 0.50 (claim); Ha : p ⫽ 0.50 z0 ⫽ ± 1.645 p⫺p ^
z⫽
⫽
0.71 ⫺ 0.50
兲共0.50兲 冪pqn 冪共0.50129
⫽
0.21 ⬇ 4.770 0.0440
Reject H0 . There is enough evidence to reject the claim. 41. Because np ⫽ 1.2 < 5, the normal distribution cannot be used to approximate the binomial distribution. 42. H0 : p ⫽ 0.34; Ha: p ⫽ 0.34 (claim) z0 ⫽ ± 2.575 p⫺p ^
z⫽
⫽
0.29 ⫺ 0.34
冪pqn 冪共0.3460兲共0.66兲
⫽
⫺0.05 ⬇ 0.820 0.061
Fail to reject H0 . There is not enough evidence to support the claim. 43. H0 : p ⫽ 0.20; Ha : p ⫽ 0.20 (claim) z0 ⫽ ± 2.575 p⫺p ^
z⫽
⫽
0.23 ⫺ 0.20
冪pqn 冪共0.2056兲共0.80兲
⫽
0.03 ⬇ 0.561 0.0534
Fail to reject H0 . There is not enough evidence to support the claim.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
44. H0 : p ⱕ 0.80 (claim); Ha : p > 0.80 z0 ⫽ 1.28 p⫺p ^
z⫽
⫽
0.85 ⫺ 0.80
冪pqn 冪共0.8043兲共0.20兲
⫽
0.05 ⬇ 0.820 0.061
Fail to reject H0 . There is not enough evidence to reject the claim. 45. H0 : p ⱕ 0.40; Ha : p > 0.40 (claim) z0 ⫽ 1.28 p⫽ ^
x 1130 ⫽ ⬇ 0.414 n 2730 p⫺p ^
z⫽
⫽
0.414 ⫺ 0.40
兲共0.60兲 冪pqn 冪共0.402730
⫽
0.14 ⬇ 1.493 0.0094
Reject H0 . There is enough evidence to support the claim. 46. H0 : p ⫽ 0.02 (claim); Ha : p ⫽ 0.02 z0 ⫽ ± 1.96 p⫽ ^
x 3 ⫽ ⬇ 0.01 n 300 p⫺p ^
z⫽
⫽
0.01 ⫺ 0.02
兲共0.98兲 冪pqn 冪共0.02300
⫽
⫺0.01 ⬇ ⫺1.24 0.0081
Fail to reject H0 . There is not enough evidence to reject the claim. 47. R2 ⫽ 30.144 48. L2 ⫽ 3.565, R2 ⫽ 29.819 49. R2 ⫽ 33.196 50. 20 ⫽ 1.145 51. H0 : 2 ≤ 2; Ha : 2 > 2 (claim)
20 ⫽ 24.769 2 ⫽
共n ⫺ 1兲s2 共17兲共2.95兲 ⫽ ⫽ 25.075 2 共2兲
Reject H0 . There is enough evidence to support the claim. 52. H0 : 2 ⱕ 60 (claim); Ha : 2 > 60
20 ⫽ 26.119 2 ⫽
共n ⫺ 1兲s2 共14兲共72.7兲 ⫽ ⬇ 16.963 2 共60兲
Fail to reject H0 . There is not enough evidence to reject the claim.
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53. H0 : 2 ⫽ 1.25 (claim); Ha : 2 ⫽ 1.25
L2 ⫽ 0.831, R2 ⫽ 12.833 2 ⫽
共n ⫺ 1兲s2 共5兲共1.03兲2 ⫽ ⬇ 3.395 2 共1.25兲2
Fail to reject H0. There is not enough evidence to reject the claim. 54. H0 : ⫽ 0.035; Ha : ⫽ 0.035 (claim)
L2 ⫽ 4.601, R2 ⫽ 32.801 2 ⫽
共n ⫺ 1兲s2 共15兲共0.026兲2 ⫽ ⬇ 8.278 2 共0.035兲2
Fail to reject H0 . There is not enough evidence to support the claim. 55. H0 : 2 ⱕ 0.01 (claim); Ha : 2 > 0.01
20 ⫽ 49.645 2 ⫽
共n ⫺ 1兲s2 共27兲共0.064兲 ⫽ ⫽ 172.800 2 共0.01兲
Reject H0. There is enough evidence to reject the claim. 56. H0 : ⱕ 0.0025 (claim); Ha : > 0.0025
20 ⫽ 27.688 2 ⫽
共n ⫺ 1兲s2 共13兲共0.0031兲2 ⫽ ⬇ 19.989 2 共0.0025兲2
Fail to reject H0 . There is not enough evidence to reject the claim.
CHAPTER 7 QUIZ SOLUTIONS 1. (a) H0: ⱖ 22 (claim); Ha: < 22 (b) “ ⱖ vs <” → Lefttailed is unknown and n ⱖ 30 → ztest. (c) z0 ⫽ ⫺2.05; Reject H0 if z < ⫺2.05. (d) z ⫽
x ⫺ 21.6 ⫺ 22 ⫺0.4 ⬇ ⫺0.580 ⫽ ⫽ s 7 0.690 冪n 冪103
(e) Fail to reject H0. There is insufficient evidence at the 2% significance level to reject the claim that the mean utilization of fresh citrus fruits by people in the U.S. is at least 22 pounds per year.
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CHAPTER 7

HYPOTHESIS TESTING WITH ONE SAMPLE
2. (a) H0: ⱖ 20 (claim); Ha: < 20 (b) “ ⱖ vs <” → Lefttailed is unknown, the population is normal, and n < 30 → ttest. (c) t0 ⫽ ⫺1.895; Reject H0 if t < ⫺1.895. (d) z ⫽
x ⫺ 18 ⫺ 20 ⫺2 ⫽ ⫽ ⬇ ⫺1.131 s 5 1.768 冪n 冪8
(e) Fail to reject H0. There is insufficient evidence at the 5% significance level to reject the claim that the mean gas mileage is at least 20 miles per gallon. 3. (a) H0 : p ⱕ 0.10 (claim); Ha : p > 0.10 (b) “ ⱕ vs >” → Righttailed np ⱖ 5 and nq ⱖ 5 → ztest (c) z0 ⫽ 1.75; Reject H0 if z > 1.75. p⫺p ^
(d) z ⫽
⫽
0.13 ⫺ 0.10
冪pqn 冪共0.1057兲共0.90兲
⫽
0.03 ⬇ 0.75 0.0397
(e) Fail to reject H0. There is insufficient evidence at the 4% significance level to reject the claim that no more than 10% of microwaves need repair during the first five years of use. 4. (a) H0 : ⫽ 113 (claim); Ha : ⫽ 113 (b) “⫽ vs ⫽ ’’ → Twotailed Assuming the scores are normally distributed and you are testing the hypothesized standard deviation → 2 test. (c) 2L ⫽ 3.565, R2 ⫽ 29.819; Reject H0 if 2 < 3.565 or if 2 > 29.819. (d) 2 ⫽
共n ⫺ 1兲s2 共13兲共108兲2 ⫽ ⬇ 11.875 2 共113兲2
(e) Fail to reject H0 . There is insufficient evidence at the 1% significance level to reject the claim that the standard deviation of the SAT critical reading scores for the state is 105. 5. (a) H0 : ⫽ $48,718 (claim); Ha : ⫽ $48,718 (b) “⫽ vs ⫽” → Twotailed is unknown, n < 30, and assuming the salaries are normally distributed → ttest. (c) not applicable (d) t ⫽
x ⫺ 47,164 ⫺ 48,718 ⫺1554 ⫽ ⫽ ⬇ ⫺0.828 s 6500 1876.388 冪n 冪12
Pvalue ⫽ 2再Area left of t ⫽ ⫺0.828冎 ⫽ 2共0.2126兲 ⫽ 0.4252 (e) Fail to reject H0. There is insufficient evidence at the 5% significance level to reject the claim that the mean annual salary for fulltime male workers ages 25 to 34 with a bachelor’s degree is $48,718.
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6. (a) H0 : ⫽ $201 (claim); Ha : ⫽ $201 (b) “⫽ vs ⫽” → Twotailed
is unknown, n ⱖ 30 → ztest. (c) not applicable (d) z ⫽
x ⫺ 216 ⫺ 201 15 ⫽ ⫽ ⬇ 2.958 s 30 5.071 冪n 冪35
Pvalue ⫽ 2再Area right of z ⫽ 2.958冎 ⫽ 2再0.0015冎 ⫽ 0.0030 (e) Reject H0 . There is sufficient evidence at the 5% significance level to reject the claim that the mean daily cost of meals and lodging for a family of four traveling in Kansas is $201.
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Hypothesis Testing with Two Samples
CHAPTER
8
8.1 TESTING THE DIFFERENCE BETWEEN MEANS (LARGE INDEPENDENT SAMPLES)
8.1 Try It Yourself Solutions Note: Answers may differ due to rounding. 1. (1) Independent (2) Dependent
2a. H0 : 1 2 ; Ha : 1 2 (claim) b. 0.01 c. z0 ± 2.575; Reject H0 if z > 2.575 or z < 2.575. d. z
共x 1 x 2 兲 共 1 2 兲
冪ns
2 1
1
s 22 n2
共3900 3500兲 共0兲 兲 冪共900 50
2
共500兲2 50
400 ⬇ 2.747 冪21200
e. Reject H0 . f. There is enough evidence to support the claim. 3a. z
共x 1 x 2 兲 共 1 2 兲
冪ns
2 1 1
s 22 n2
共293 286兲 共0兲 24兲 共18兲 冪共150 200 2
2
7 ⬇ 3.00 冪5.46
→ Pvalue 再area right of z 3.00冎 0.0014 b. Reject H0 . There is enough evidence to support the claim.
8.1 EXERCISE SOLUTIONS 1. Two samples are dependent if each member of one sample corresponds to a member of the other sample. Example: The weights of 22 people before starting an exercise program and the weights of the same 22 people 6 weeks after starting the exercise program. Two samples are independent if the sample selected from one population is not related to the sample from the second population. Example: The weights of 25 cats and the weights of 25 dogs. 2. State the hypotheses and identify the claim. Specify the level of significance and find the critical value(s). Identify the rejection regions. Find the standardized test statistic. Make a decision and interpret in the context of the claim. 3. Use Pvalues. 4. (1) The samples must be randomly selected. (2) The samples must be independent. (3) 再n1 30 and n2 30冎 or 再each population must be normally distributed with known standard deviations冎
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5. Independent because different students were sampled. 6. Dependent because the same students were sampled. 7. Dependent because the same adults were sampled. 8. Independent because different individuals were sampled. 9. Independent because different boats were sampled. 10. Dependent because the same cars were sampled. 11. Dependent because the same tire sets were sampled. 12. Dependent because the same people were sampled. 13. H0 : 1 2 (claim); Ha : 1 2 Rejection regions: z0 < 1.96 and z0 > 1.96 (Twotailed test) (a) x1 x2 16 14 2 (b) z
共x 1 x 2 兲 共 1 2 兲
冪
共16 14兲 共0兲
冪
共1.1兲 共1.5兲 n1 n2 50 50 (c) z is in the rejection region because 7.60 > 1.96. s 21
s 22
2
2
2 冪0.0692
⬇ 7.60
(d) Reject H0. There is enough evidence to reject the claim. 14. H0: 1 2 ; H1: 1 > 2 (claim) Rejection region: z0 > 1.28 (Righttailed test) (a) x1 x2 500 510 10 (b) z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s22 n2
共500 510兲 共0兲
30兲 共15兲 冪共100 75 2
2
10 ⬇ 2.89 冪12
(c) z is not in the rejection region because 2.89 < 1.28. (d) Fail to reject H0. There is not enough evidence to support the claim. 15. H0: 1 2 ; Ha : 1 < 2 (claim) Rejection region: z0 < 2.33 (Lefttailed test) (a) x1 x2 1225 1195 30 (b) z
共x 1 x 2 兲 共 1 2 兲
冪n
s 21 1
s 22 n2
共1225 1195兲 共0兲
冪
共75兲 共105兲 35 105 2
2
30 冪265.714
⬇ 1.84
(c) z is not in the rejection region because 1.84 > 2.330. (d) Fail to reject H0. There is not enough evidence to support the claim.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
16. H0: 1 2 (claim); H1: 1 > 2 Rejection region: z0 > 1.88 (Righttailed test) (a) x1 x2 5004 4895 109 (b) z
共x1 x2 兲 共 1 2兲
冪
s21 n1
s22
共5004 4895兲 共0兲
冪
n2
共136兲 共215兲 144 156 2
2
109 冪424.759
⬇ 5.29
(c) z is in the rejection region because 5.29 > 1.88. (d) Reject H0. There is enough evidence to reject the claim. 17. H0 : 1 2; Ha : 1 > 2 (claim) z0 2.33; Reject H0 if z > 2.33. z
共x 1 x 2 兲 共 1 2 兲
冪n
s 21
s 22
n2
1
共5.2 5.5兲 共0兲
冪共0.245兲
2
共0.3兲2 37
.30 ⬇ 5.207 冪0.00332
Fail to reject H0. There is not enough evidence to support the claim. 18. H0: 1 2 (claim); Ha: 1 2 z0 ± 1.96 z
共x1 x2 兲 共 1 2 兲
冪n
s 21
s 22
n2
1
共52 45兲 共0兲
冪
共2.5兲 共5.5兲 70 60 2
2
7
冪0.59345
⬇ 9.087
Reject H0. There is enough evidence to support the claim. 19. (a) H0 : 1 2 ; Ha: 1 2 (claim) (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. (c) z
共x 1 x 2 兲 共 1 2 兲
冪ns
2 1
1
s 22 n2
共42 45兲 共0兲
冪共4.735兲
2
共4.3兲2 35
3 ⬇ 2.786 冪1.159
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to support the claim that the mean braking distance is different for both types of tires. 20. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) z0 1.28; Reject H0 if z > 1.28. (c) z
共x1 x2 兲 共 1 2 兲
冪n
s 21 1
s 22 n2
共55 51兲 共0兲
冪共5.350兲
2
共4.9兲 50
2
4 ⬇ 3.92 冪1.042
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to support the claim that the mean braking distance for Type C is greater than for Type D.
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21. (a) H0: 1 2 ; Ha : 1 < 2 (claim) (b) z0 2.33; Reject H0 if z < 2.33. (c) z
共x 1 x 2 兲 共 1 2 兲
冪n
s 21
1
s 22
n2
共75 80兲 共0兲
冪
共12.50兲 共20兲 47 55 2
2
5 冪10.597
⬇ 1.54
(d) Fail to reject H0. (e) There is insufficient evidence at the 1% significance level to conclude that the repair costs for Model A are lower than for Model B. 22. (a) H0: 1 2 (claim); Ha: 1 2 (b) z0 ± 2.575; Reject H0 if z < 2.575 or z > 2.575. (c) z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s 22
n2
共50 60兲 共0兲
冪
共10兲2 共18兲2 34 46
10 冪9.985
⬇ 3.165
(d) Reject H0. (e) There is sufficient evidence at the 1% significance level to reject the claim that the mean repair costs for Model A and Model B are the same. 23. (a) H0: 1 2 (claim); Ha : 1 2 (b) z0 ± 2.575; Reject H0 if z < 2.575 or z > 2.575. (c) z
共x 1 x 2 兲 共 1 2 兲
冪n
s 21
1
s22
n2
共21.0 20.8兲 共0兲
冪
共5.0兲 共4.7兲 43 56 2
2
0.2 冪0.976
⬇ 0.202
(d) Fail to reject H0. (e) There is insufficient evidence at the 1% significance level to reject the claim that the male and female high school students have equal ACT scores. 24. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) z0 1.28; Reject H0 if z > 1.28. (c) z
共x1 x2 兲 共 1 2兲
冪n
s 21
1
s 22
n2
共22.2 20.0兲 共0兲
冪共4.849兲
2
共5.4兲2 4
2.2 ⬇ 2.07 冪1.133
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to support the claim that the ACT scores are higher for high school students in a college prep program. 25. (a) H0: 1 2 (claim); Ha : 1 2 (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. (c) z
共x 1 x 2 兲 共 1 2 兲
冪n
s 21 1
s 22 n2
共131 136兲 共0兲
冪
共26兲 共19兲 35 35 2
2
5 冪29.629
⬇ 0.919
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
(d) Fail to reject H0. (e) There is insufficient evidence at the 10% significance level to reject the claim that the lodging cost for a family traveling in North Carolina is the same as South Carolina. 26. (a) H0: 1 2 ; Ha: 1 < 2 (claim) (b) z0 1.645; Reject H0 if z < 1.645. (c) z
共x1 x2 兲 共 1 2兲
冪n
s 21
1
s 22
n2
共2015 2715兲 共0兲
冪
共113兲 共97兲 30 30 2
2
700 冪739.267
⬇ 25.745
(d) Reject H0. (e) There is sufficient evidence at the 5% significance level to support the claim that households in the United States headed by people under the age of 25 spend less on food away from home than households headed by people ages 55–64. 27. (a) H0: 1 2 (claim); Ha : 1 2 (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. (c) z
共x1 x 2 兲 共 1 2 兲
冪n
s 21
1
s 22
n2
共145 138兲 共0兲
冪
共28兲 共24兲 50 50 2
2
7 冪27.2
⬇ 1.342
(d) Fail to reject H0. (e) There is insufficient evidence at the 10% significance level to reject the claim that the lodging cost for a family traveling in North Carolina is the same as South Carolina. The new samples do not lead to a different conclusion. 28. H0: 1 1; Ha: 1 < 2 (claim) z0 1.645; Reject H0 if z < 1.645. z
共x1 x2 兲 共 1 2 兲
冪n
s 21 1
s 22 n2
共2130 2655兲 共0兲
冪共12440 兲
2
共116兲 40
2
525 19.555 冪720.8
Reject H0. There is enough evidence to support the claim. 29. (a) H0: 1 2 ; Ha: 1 < 2 (claim) (b) z0 1.96; Reject H0 if z > 1.96. (c) x1 ⬇ 2.130, s1 ⬇ 0.490, n1 30 x2 ⬇ 1.757, s2 ⬇ 0.470, n2 30 z
共x1 x2兲 共1 2兲
冪n
s 21 1
s 22 n2
共2.130 1.757兲 共0兲 兲 冪共0.490 30
2
共0.470兲2 30
0.373 ⬇ 3.01 冪0.0154
(d) Reject H0. (e) At the 2.5% level of significance, there is sufficient evidence to support the claim. 30. (a) H0: 1 2 ; Ha: 1 < 2 (claim) (b) z0 1.88; Reject H0 if z < 1.88.
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(c) x1 ⬇ 32.523, s1 ⬇ 4.477, n1 35 x2 47.989, s2 ⬇ 4.651, n2 35 z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s 22
n2
共32.523 47.989兲 共0兲 兲 冪共4.477 35
2
共4.651兲 35
2
15.466 ⬇ 14.17 冪1.191
(d) Reject H0. (e) There is sufficient evidence at the 3% significance level to support the sociologist’s claim. 31. (a) H0: 1 2 (claim); Ha: 1 2 (b) z0 ± 2.575; Reject H0 if z < 2.575 or z > 2.575. (c) x1 ⬇ 0.875, s1 ⬇ 0.011, n1 35 x2 ⬇ 0.701, s2 ⬇ 0.011, n2 35 z
共x 1 x 2 兲 共 1 2 兲
冪ns
2 1
1
s 22 n2
共0.875 0.701兲 共0兲 兲 冪共0.011 35
2
共0.011兲2 35
0.174 ⬇ 66.172 冪0.0000006914
(d) Reject H0. (e) At the 1% level of significance, there is sufficient evidence to reject the claim. 32. (a) H0: 1 2 (claim); Ha: 1 2 (b) z0 ⬇ ± 2.05; Reject H0 if z < 2.05 or z > 2.05. (c) x1 3.337, s1 0.011, n1 40 x2 3.500, s2 0.010, n2 40 z
共x1 x2 兲 共 1 2 兲
冪
s21 n1
s22 n2
共3.337 3.500兲 共0兲
冪
共0.011兲 共0.010兲 40 40 2
2
0.163 冪0.000005525
⬇ 69.346
(d) Reject H0. (e) There is sufficient evidence at the 4% level of significance to reject the claim. 33. They are equivalent through algebraic manipulation of the equation.
1 2 → 1 2 0 34. They are equivalent through algebraic manipulation of the equation.
1 2 → 1 2 0 35. H0: 1 2 9 (claim); Ha: 1 2 9 z0 ± 2.575; Reject H0 if z < 2.575 or z > 2.575. z
共x 1 x 2 兲 共 1 2 兲
冪ns
2 1 1
s 22 n2
共11.5 20兲 共9兲
冪共3.870兲
2
共6.7兲2 65
0.5 ⬇ 0.528 冪0.897
Fail to reject H0. There is not enough evidence to reject the claim.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
36. H0: 1 2 2 (claim); Ha: 1 2 2 z0 ± 1.96; Reject H0 if z < 1.96 or z > 1.96. z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s 22
共12.95 15.02兲 共2兲
冪
n2
共4.31兲 共4.99兲 48 56 2
2
0.07 冪0.8316
⬇ 0.077
Fail to reject H0. There is not enough evidence to reject the claim. 37. H0: 1 2 6000; Ha: 1 2 > 6000 (claim) z0 1.28; Reject H0 if z > 1.28. z
共x 1 x 2 兲 共 1 2 兲
冪ns
2 1
1
s 22 n2
共67,900 64,000兲 共6000兲
兲 冪共8875 45
2
共9175兲2 42
2100 ⬇ 1.084 冪3,754,647.817
Fail to reject H0. There is not enough evidence to support the claim. 38. H0: 1 2 30,000; Ha: 1 2 > 30,000 (claim) z0 1.645; Reject H0 if z > 1.645. z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s 22
共54,900 27,200兲 共30,000兲
冪
n2
共8250兲 共3200兲 31 33 2
2
2300 冪2,505,867.546
1.453
Fail to reject H0. There is not enough evidence to support the claim.
冪ns
2 1
39. 共x1 x2兲 zc
1
冪ns
s 22 < 1 2 < 共x1 x2兲 zc n2
兲 冪共9.9 269
2
共123.1 125兲 1.96
2 1 1
s22 n2
共10.1兲2 < 1 2 < 共123.1 125兲 1.96 268
兲 冪共9.9 269
2
共10.1兲2 268
1.9 1.96冪0.745 < 1 2 < 1.9 1.96冪0.745 3.6 < 1 2 < 0.2
冪ns
2 1
共x1 x2兲 zc
40.
1
冪
共10.3 8.5兲 1.96
冪ns
s 22 < 1 2 < 共x1 x2兲 zc n2
共1.2兲 共1.5兲 < 1 2 < 共10.3 8.5兲 1.96 140 127 2
2
2 1
1
冪
s 22 n2
共1.2兲2 共1.5兲2 140 127
1.8 1.96冪0.028 < 1 2 < 1.8 1.96冪0.028 1.5 < 1 2 < 2.1 41. H0: 1 2 0; Ha: 1 2 < 0 (claim) z0 1.645; Reject H0 if z < 1.645. z
共x 1 x 2兲 共 1 2兲
冪ns
2 1 1
s 22 n2
共123.1 125兲 共0兲 兲 冪共9.9 269
2
共10.1兲2 268
1.9 ⬇ 2.20 冪0.745
Reject H0. There is enough evidence to support the claim. I would recommend using the DASH diet and exercise program over the traditional diet and exercise program.
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42. H0: 1 2 0; Ha: 1 2 > 0 (claim) z0 1.645; Reject H0 if z > 1.645. z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s 22
n2
共10.3 8.5兲 共0兲
冪
共1.2兲 共1.5兲 140 127 2
2
1.8 冪0.028
⬇ 10.76
Reject H0. There is enough evidence to support the claim. I would recommend using Irinotecan over Fluorouracil because the average number of months with no reported cancer related pain was significantly higher. 43. H0: 1 2 0; Ha: 1 2 < 0 (claim) The 95% CI for 1 2 in Exercise 39 contained values greater than or equal to zero and, as found in Exercise 41, there was enough evidence at the 5% level of significance to support the claim. If zero is not contained in the CI for 1 2, you reject H0 because the null hypothesis states that 1 2 is greater than or equal to zero. 44. H0: 1 2 0; H1: 1 2 > 0 (claim) The 95% CI for 1 2 in Exercise 40 contained only values greater than zero and, as found in Exercise 42, there was sufficient evidence at the 5% level of significance to support the claim. If the CI for 1 2 contains only positive numbers, you reject H0 , because the null hypothesis states that 1 2 is less than or equal to zero.
8.2 TESTING THE DIFFERENCE BETWEEN MEANS (SMALL INDEPENDENT SAMPLES)
8.2 Try It Yourself Solutions 1a. H0 : 1 2 ; Ha: 1 2 (claim) b. 0.05 c. d.f. min再n1 1, n2 1冎 min再8 1, 10 1冎 7 d. t0 ± 2.365; Reject H0 if t < 2.365 or t > 2.365. e. t
共x 1 x 2 兲 共 1 2 兲
冪n
s21 1
s22 n2
共141 151兲 共0兲
冪共7.08 兲
2
共3.1兲2 10
10 ⬇ 3.757 冪7.086
f. Reject H0. g. There is enough evidence to support the claim. 2a. H0 : 1 2 ; Ha: 1 < 2 (claim)
b. 0.10
c. d.f. n1 n2 2 12 15 2 25 e. t
d. t0 1.316; Reject H0 if t < 1.316.
共x1 x2 兲 共 1 2 兲
冪共n
1兲s 21 共n2 1兲s 22 n1 n2 2 3 ⬇ 3.997 冪3.755 冪0.15 1
共32 35兲 共0兲
冪n1 n1 冪共12 1兲共122.1兲 15共15 2 1兲共1.8兲 冪121 151 1
2
2
2
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
f. Reject H0 . g. There is enough evidence to support the claim.
8.2 EXERCISE SOLUTIONS 1. State hypotheses and identify the claim. Specify the level of significance. Determine the degrees of freedom. Find the critical value(s) and identify the rejection region(s). Find the standardized test statistic. Make a decision and interpret in the context of the original claim. 2. (1) The samples must be randomly selected. (2) The samples must be independent. (3) Each population must have a normal distribution. 3. (a) d.f. n1 n2 2 23 t0 ± 1.714
4. (a) d.f. n1 n2 2 25 t0 2.485
(b) d.f. min再n1 1, n2 1冎 10 t0 ± 1.812
(b) d.f. min再n1 1, n2 1冎 11 t0 2.718
5. (a) d.f. n1 n2 2 22 t0 2.074
6. (a) d.f. n1 n2 2 39 t0 ± 1.96
(b) d.f. min再n1 1, n2 1冎 8 t0 2.306
(b) d.f. min再n1 1, n2 1冎 18 t0 ± 2.101
7. (a) d.f. n1 n2 2 19
8. (a) d.f. n1 n2 2 11
t0 1.729
t0 1.363
(b) d.f. min再n1 1, n2 1冎 7
(b) d.f. min再n1 1, n2 1冎 3
t0 1.895
t0 1.638
9. (a) d.f. n1 n2 2 27
10. (a) d.f. n1 n2 2 16
t0 ± 2.771
t0 2.921
(b) d.f. min再n1 1, n2 1冎 11
(b) d.f. min再n1 1, n2 1冎 6
t0 ± 3.106
t0 3.707
11. H0: 1 2 (claim); Ha: 1 2 d.f. n1 n2 2 15 t0 ± 2.947 (Twotailed test) (a) x1 x2 33.7 35.5 1.8 (b) t
共x1 x 2 兲 共 1 2 兲
冪共n
1兲s 21 共n2 1兲s 22 n1 n2 2 1.8 ⬇ 1.199 冪9.286 冪0.243 1
共33.7 35.5兲 共0兲
冪n1 n1 冪共10 1兲共103.5兲 7 共7 2 1兲共2.2兲 冪101 71 1
2
2
2
(c) t is not in the rejection region. (d) Fail to reject H0.
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12. H0: 1 2 (claim); Ha: 1 < 2 d.f. n1 n2 2 18 t0 1.330 共Lefttailed test兲 (a) x1 x2 0.515 0.475 0.04 (b) t
共x1 x2兲 共 1 2兲
冪共n
1
1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共0.515 0.475兲 共0兲 兲 共9 1兲共0.215兲 冪共11 1兲共0.305 冪111 91 11 9 2 2
2
0.4 ⬇ 0.331 冪0.0722 冪0.202
(c) t is not in the rejection region. (d) Fail to reject H0. 13. There is no need to run the test since this is a righttailed test and the test statistic is negative. It is obvious that the standardized test statistic will also be negative and fall outside of the rejection region. So, the decision is to fail to reject H0. 14. H0: 1 2 (claim); Ha: 1 > 2 d.f. min再n1 1, n2 1冎 13 (a) x1 x2 45 50 5 (b) t0 2.650 (Righttailed test) t
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s22 n2
共45 50兲 共0兲
冪
共4.8兲 共1.2兲 16 14 2
2
5 冪1.543
⬇ 4.025
(c) t is not in the rejection region. (d) Fail to reject H0. 15. (a) H0: 1 2 (claim); Ha: 1 2 (b) d.f. n1 n2 2 12 17 2 27 t0 ± 1.703; Reject H0 if t < 1.703 or t > 1.703. (c) t
共x1 x2兲 共 1 2兲
冪共n
1
1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共10.1 8.3兲 共0兲
冪
共12 1兲共4.11兲 共17 1兲共4.02兲 12 17 2 (d) Fail to reject H0. 2
2
冪
1 1 12 17
1.8 冪16.459 冪0.142
⬇ 1.177
(e) There is not enough evidence to reject the claim. 16. (a) H0: 1 2 (claim); Ha: 1 2 (b) d.f. n1 n2 2 5 8 2 11 t0 ± 2.201; Reject H0 if t < 2.201 or t > 2.201. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 8
(c) t
HYPOTHESIS TESTING WITH TWO SAMPLES
共x1 x2兲 共 1 2兲
冪共n
1

1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共11.0 10.6兲 共0兲
冪共5 1兲共4.075 兲 8 共8 2 1兲共6.62兲 冪15 81 2
2
0.4 ⬇ 0.120 冪33.912 冪0.325
(d) Fail to reject H0. (e) There is not enough evidence to reject the claim. 17. (a) H0: 1 2 ; Ha: 1 < 2 (claim) (b) d.f. n1 n2 2 35 t0 1.282; Reject H0 if t < 1.282. (c) t
共x1 x2兲 共 1 2兲
冪共n
1
兲 共23 1兲共205兲 冪n1 n1 冪共14 1兲共190 冪141 231 14 23 2
1兲s21 共n2 1兲s22 n1 n2 2 268
冪39,824.286 冪0.115
共473 741兲 共0兲
1
2
2
2
3.960
(d) Reject H0. (e) There is enough evidence to support the claim. 18. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) d.f. n1 n2 2 11 t0 1.363; Reject H0 if t > 1.363. (c) t
共x1 x2兲 共 1 2兲
冪
共n1 1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共1090 485兲 共0兲
冪共5 1兲共4035 兲 8 共8 2 1兲共382兲 冪15 81 2
2
605 ⬇ 2.723 冪151,918.546 冪0.325
(d) Reject H0. (e) There is enough evidence to support the claim. 19. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) d.f. min再n1 1, n2 1冎 14 t0 1.345; Reject H0 if t > 1.345. (c) z
共x 1 x 2兲 共 1 2兲
冪ns
2 1 1
s22 n2
共42,200 37,900兲 共0兲 兲 冪共8600 19
2
共5500兲2 15
4300 ⬇ 1.769 冪5,909,298.246
(d) Reject H0. (e) There is enough evidence to support the claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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HYPOTHESIS TESTING WITH TWO SAMPLES
20. (a) H0: 1 2; Ha: 1 2 (claim) (b) d.f. n1 n2 2 33 t0 ± 2.576; Reject H0 if t < 2.576 or t > 2.576. (c) t
共x1 x2兲 共 1 2兲
冪共n
1
1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共36,700 34,700兲 共0兲
冪
共17 1兲共7800兲2 共18 1兲共7375兲2 17 18 2 2000
冪57,517,594.70 冪0.1144
冪171 181
⬇ 0.780
(d) Fail to reject H0. (e) There is not enough evidence to support the claim. 21. (a) H0: 1 2 ; Ha: 1 2 (claim) (b) d.f. n1 n2 2 21 t0 ± 2.831; Reject H0 if t < 2.831 or t > 2.831. (c) x1 340.300, s1 22.301, n1 10 x2 389.538, s2 14.512, n2 13 共x1 x2兲 共 1 2 兲 t 共n1 1兲s21 共n2 1兲s22 1 1 n1 n2 2 n1 n2
冪
冪
共340.300 389.538兲 共0兲
冪
共10 1兲共22.301兲 共13 1兲共14.512兲 10 13 2 (d) Reject H0. 2
2
冪
1 1 10 13
49.238 冪333.485 冪0.177
⬇ 6.410
(e) There is enough evidence to support the claim. 22. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) d.f. min再n1 1, n2 1冎 13 t0 1.350; Reject H0 if t > 1.350. (c) x1 402.765, s1 11.344, n1 17 x2 384.000, s2 17.698, n2 14 t
共x1 x2 兲 共 1 2 兲
冪ns
2 1 1
s 22 n2
共402.765 384.000兲 共0兲 兲 冪共11.344 17
2
共17.698兲2 14
18.765 ⬇ 3.429 冪29.943
(d) Reject H0. (e) There is enough evidence to support the claim and to recommend using the experimental method.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
23. (a) H0: 1 2 ; Ha: 1 < 2 (claim) (b) d.f. n1 n2 2 42 t0 1.282 → t < 1.282 (c) x1 56.684, s1 6.961, n1 19 x2 67.400, s2 9.014, n2 25 t
共x1 x2兲 共 1 2 兲
冪共n
1
1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共56.684 67.400兲 共0兲 兲 共25 1兲共9.014兲 1 1 冪共19 1兲共6.961 冪 19 25 2 19 25 2
2
10.716 ⬇ 4.295 冪67.196 冪0.0926
(d) Reject H0. (e) There is enough evidence to support the claim and to recommend changing to the new method. 24. (a) H0: 1 2 ; Ha: 1 < 2 (claim) (b) d.f. n1 n2 2 39 t0 1.645; Reject H0 if t < 1.645. (c) x1 79.091, s1 6.900, n1 22 x2 83.000, s2 7.645, n2 19 t
共x1 x2兲 共 1 2兲
冪共n
1
1兲s 21 共n2 1兲s 22 n1 n2 2
冪n1 n1 1
2
共79.091 83.000兲 共0兲 兲 共19 1兲共7.645兲 1 1 冪共22 1兲共6.900 冪 22 19 2 22 19 2
2
3.909 ⬇ 1.721 冪52.611 冪0.0981
(d) Reject H0. (e) There is enough evidence to support the claim. 25. ^
冪共n
1
1兲s 21 共n2 1兲s22 n1 n2 2
共x1 x2兲 ± tc
^
冪n1 n1 1
冪共15 1兲共156.2兲 12共12 2 1兲共8.1兲 2
→ 共450 420兲 ± 2.060
2
2
⬇ 7.099
冪151 121
7.099
→ 30 ± 5.664 → 24.336 < 1 2 < 35.664 → 24 < 1 2 < 36 26. ^
冪
共n1 1兲s 21 共n2 1兲s 22 n1 n2 2
共x1 x2兲 ± tc
^
冪共15 1兲共152.1兲 12共12 2 1兲共1.8兲 2
2
⬇ 1.974
冪n1 n1 → 共37 32兲 ± 2.060 1.974冪151 121 1
2
→ 5 ± 1.575 → 3.425 < 1 2 < 6.575 → 3 < 1 2 < 7
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227
228
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HYPOTHESIS TESTING WITH TWO SAMPLES
冪ns
2 1
27. 共x1 x2兲 ± tc
1
s22 → 共75 70兲 ± 1.771 n2
兲 冪共3.64 16
2
共2.12兲2 14
→ 5 ± 1.898 → 3.102 < 1 2 < 6.898 → 3 < 1 2 < 7
冪
28. 共x1 x2兲 ± tc
s21 s2 2 → 共39 37兲 ± 1.345 n1 n2
冪
共2.42兲2 共1.65兲2 20 12
→ 2 ± 0.926 → 1.074 < 1 2 < 2.926 → 1 < 1 2 < 3
8.3 TESTING THE DIFFERENCE BETWEEN MEANS (DEPENDENT SAMPLES)
8.3 Try It Yourself Solutions 1.
2.
Before
After
d
d2
72 81 76 74 75 80 68 75 78 76 74 77
73 80 79 76 76 80 74 77 75 74 76 78
1 1 3 2 1 0 6 2 3 2 2 1
1 1 9 4 1 0 36 4 9 4 4 1
兺d 12
兺d 2 74
Before
After
d
d2
101.8 98.5 98.1 99.4 98.9 100.2 97.9
99.2 98.4 98.2 99 98.6 99.7 97.8
2.6 0.1 0.1
0.1
6.76 0.01 0.01 0.16 0.09 0.25 0.01
兺d 3.9
兺d 2 7.29
0.4 0.3 0.5
a. H0 : d 0; Ha: d < 0 (claim) b. 0.05 and d.f. n 1 11 c. t0 ⬇ 1.796; Reject H0 if t < 1.796. 兺d 12 1 d. d n 12 sd
兲 冪n共兺dn共n兲 共1兺d兲 兲 冪12共7412兲 共11共12 兲 2
2
2
⬇ 2.374
d d 1 0 ⬇ 1.459 sd 2.374 冪n 冪12 f. Fail to reject H0. e. t
g. There is not enough evidence to support the claim.
a. H0 : d 0; Ha: d 0 (claim) b. 0.05 and d.f. n 1 6 c. t0 ± 2.447; Reject H0 if t < 2.447 or t > 2.447. d. d sd
兺d 3.9 ⬇ 0.557 n 7
冪n共兺dn共n兲 共1兺d兲 兲 冪7共7.297兲共6兲 共3.9兲 2
2
2
⬇ 0.924
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CHAPTER 8
e. t

HYPOTHESIS TESTING WITH TWO SAMPLES
d d 0.557 0 ⬇ 1.595 sd 0.924 冪n 冪7
f. Fail to reject H0. g. There is not enough evidence at the 5% significance level to conclude that the drug changes the body’s temperature.
8.3 EXERCISE SOLUTIONS 1. (1) Each sample must be randomly selected from a normal population. (2) Each member of the first sample must be paired with a member of the second sample. 2. The symbol d represents the mean of the differences between the paired data entries in dependent samples. The symbol Sd represents the standard deviation of the differences between the paired data entries in the dependent samples. 3. H0: d 0; Ha: d < 0 (claim)
0.05 and d.f. n 1 13 t0 1.771 (Lefttailed) t
d d 1.5 0 1.5 ⬇ 1.754 sd 3.2 855 冪n 冪14
Fail to reject H0. 4. H0: d 0 (claim); Ha: d 0
0.01 and d.f. n 1 7 t0 ± 3.499 (Twotailed) t
d d 3.2 0 3.2 ⬇ 1.071 sd 8.45 2.988 冪n 冪8
Fail to reject H0. 5. H0: d 0 (claim); Ha: d > 0
0.10 and d.f. n 1 15 t0 1.341 (Righttailed) t
d d 6.5 0 6.5 ⬇ 2.725 sd 9.54 2.385 冪n 冪16
Reject H0. 6. H0: d 0; Ha: d > 0 (claim)
0.05 and d.f. n 1 27 t0 1.703 (Righttailed)
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229
230
t
CHAPTER 8

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0.55 d d 0.55 0 ⬇ 2.940 sd 0.99 0.187 冪n 冪28
Reject H0. 7. H0: d 0 (claim); Ha: d < 0
0.01 and d.f. n 1 14 t0 2.624 共 Lefttailed兲 t
d d 2.3 0 2.3 ⬇ 7.423 sd 1.2 0.3098 冪n 冪15
Reject H0. 8. H0: d 0; Ha: d 0 (claim)
0.10 and d.f. n 1 19 t0 ± 1.729 (Twotailed) t
1 d d 1 0 ⬇ 1.626 sd 2.75 0.615 冪n 冪20
Fail to reject H0. 9. (a) H0: d 0; Ha: d < 0 (claim) (b) t0 2.650; Reject H0 if t < 2.650. (c) d ⬇ 33.714 and sd ⬇ 42.034 (d) t
d d 33.714 0 33.714 ⬇ 3.001 sd 42.034 11.234 冪n 冪14
(e) Reject H0. (f) There is enough evidence to support the claim that the SAT scores improved. 10. (a) H0: d 0; Ha: d < 0 (claim) (b) t0 2.821; Reject H0 if t < 2.821. (c) d ⬇ 59.9 and sd ⬇ 26.831 (d) t
d d 59.9 0 59.9 ⬇ 7.060 sd 26.831 8.485 冪n 冪10
(e) Reject H0. (f) There is enough evidence to support the claim that the SAT prep course improves verbal SAT scores. 11. (a) H0: d 0; Ha: d < 0 (claim) (b) t0 1.415; Reject H0 if t < 1.415. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
(c) d ⬇ 1.575 and sd ⬇ 0.803 (d) t
d d 1.575 0 1.575 5.548 sd 0.803 0.284 冪n 冪8
(e) Reject H0. (f) There is enough evidence to support the fuel additive improved gas mileage. 12. (a) H0: d 0; Ha: d < 0 (claim) (b) t0 1.397; Reject H0 if t < 1.397. (c) d ⬇ 1.567 and sd ⬇ 0.760 (d) t
d d 1.567 0 1.567 ⬇ 6.186 sd 0.760 0.253 冪n 冪9
(e) Reject H0. (f) There is enough evidence to support the claim that the fuel additive improved gas mileage. 13. (a) H0: d 0; Ha: > 0 (claim) (b) t0 1.363; Reject H0 if t > 1.363. (c) d 3.75 and sd ⬇ 7.84 (d) t
d d 3.75 0 3.75 1.657 sd 7.84 2.26 冪n 冪12
(e) Reject H0. (f) There is enough evidence to support the claim that the exercise program helps participants lose weight. 14. (a) H0: d 0; Ha: d > 0 (claim) (b) t0 1.350; Reject H0 if t > 1.350. (c) d 1.357 and sd 3.97 (d) t
d d 1.357 0 1.357 1.278 sd 3.97 1.06 冪n 冪14
(e) Fail to reject H0. (f) There is not enough evidence to support the claim that the program helped adults lose weight. 15. (a) H0: d 0; Ha: d > 0 (claim) (b) t0 2.764; Reject H0 if t > 2.764. (c) d ⬇ 1.255 and sd ⬇ 0.4414
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(d) t

HYPOTHESIS TESTING WITH TWO SAMPLES
d d 1.255 0 1.255 ⬇ 9.438 sd 0.4414 0.1331 冪n 冪11
(e) Reject H0. (f) There is enough evidence to support the claim that soft tissue therapy and spinal manipulation help reduce the length of time patients suffer from headaches. 16. (a) H0: d 0; Ha: d < 0 (claim) (b) t0 1.761; Reject H0 if t < 1.761. (c) d ⬇ 48.467 and sd ⬇ 239.005 (d) t
d d 48.467 0 48.467 ⬇ 0.785 sd 239.005 61.711 冪n 冪15
(e) Fail to reject H0. (f) There is not enough evidence to support the claim that Vitamin C will increase muscular endurance. 17. (a) H0: d 0; Ha: d > 0 (claim) (b) t0 1.895; Reject H0 if t > 1.895. (c) d 14.75 and sd ⬇ 6.861 (d) t
d d 14.75 0 14.75 ⬇ 6.081 sd 6.86 2.4257 冪n 冪8
(e) Reject H0. (f) There is enough evidence to support the claim that the new drug reduces systolic blood pressure. 18. (a) H0: d 0; Ha: d > 0 (claim) (b) t0 1.860; Reject H0 if t > 1.860. (c) d ⬇ 6.333 and sd ⬇ 5.362 (d) t
d d 6.333 0 6.333 ⬇ 3.544 sd 5.362 1.787 冪n 冪9
(e) Reject H0. (f) There is enough evidence to support the claim that the new drug reduces diastolic blood pressure. 19. (a) H0: d 0; Ha: d 0 (claim) (b) t0 ± 2.365; Reject H0 if t < 2.365 or t > 2.365. (c) d 1 and sd 1.309
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
1 d d 1 0 2.160 sd 1.31 0.463 冪n 冪8
(d) t
(e) Fail to reject H0. (f) There is not enough evidence to support the claim that the product ratings have changed. 20. (a) H0: d 0; Ha: d 0 (claim) (b) t0 ± 2.571; Reject H0 if t < 2.571 or t > 2.571. (c) d ⬇ 7.167 and sd ⬇ 10.245 x d 7.167 0 7.167 ⬇ 1.713 sd 10.245 4.183 冪n 冪6
(d) t
(e) Fail to reject H0. (f) There is not enough evidence to support the claim that the plant performance ratings have changed. 21. d ⬇ 1.525 and sd ⬇ 0.542 d t兾2
sd 冪n
< d < d t兾2
1.525 1.753
冢0.542 冣< 冪16
sd 冪n
d
< 1.525 1.753
冢0.542 冣 冪16
1.525 0.238 < d < 1.525 0.238 1.76 < d < 1.29 22. d 0.436 and sd ⬇ 0.677 d t兾2 0.436 2.160
sd 冪n
< d < d t兾2
< 冢0.677 冪14 冣
d
sd 冪n
< 0.436 2.160
冢0.677 冪14 冣
0.436 0.391 < d < 0.436 0.391 0.83 < d < 0.05
8.4 TESTING THE DIFFERENCE BETWEEN PROPORTIONS
8.4 Try It Yourself Solutions 1a. H0 : p1 p2 ; Ha: p1 p2 (claim)
b. 0.05
c. z0 ± 1.96; Reject H0 if z < 1.96 or z > 1.96. d. p
x1 x2 1484 1497 0.217 n1 n2 6869 6869
q 0.783 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
233
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e. n1 p ⬇ 1490.573 > 5, n1 q ⬇ 5378.427 > 5, n2 p ⬇ 1490.573 > 5, and n2 q ⬇ 5378.427 > 5.
共 p1 p2兲 共 p1 p2 兲 ^
f. z
共0.216 0.218兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.217 0.783 6869 6869
冣
0.002 冪0.00004947
⬇ 0.284
g. Fail to reject H0 . h. There is not enough evidence to support the claim. b. 0.05
2a. H0 : p1 p2 ; Ha: p1 > p2 (claim) c. z0 1.645; Reject H0 if z > 1.645. x1 x2 1264 522 0.130 n1 n2 6869 6869
d. p
q 0.870 e. n1 p ⬇ 892.97 > 5, n1 q ⬇ 5976.03 > 5, n2 p ⬇ 892.97 > 5, and n2 q ⬇ 5976.03 > 5.
共 p1 p2 兲 共 p1 p2 兲 ^
f. z
共0.184 0.076兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.130 0.870 6869 6869
冣
0.108 冪0.00003293
⬇ 18.820
g. Reject H0 . h. There is enough evidence to support the claim.
8.4 EXERCISE SOLUTIONS 1. State the hypotheses and identify the claim. Specify the level of significance. Find the critical value(s) and rejection region(s). Find p and q. Find the standardized test statistic. Make a decision and interpret in the context of the claim. 2. (1) The samples must be randomly selected. (2) The sample must be independent. (3) n1 p1 5, n1q1 5, n2 p2 5, and n2q2 5 3. H0: p1 p2 ; Ha: p1 p2 (claim) z0 ± 2.575 (Twotailed test) x1 x2 35 36 0.546 n1 n2 70 60 q 0.454 共 p p2兲 共 p1 p2 兲 共0.500 0.600兲 共0兲 0.100 z 1 ⬇ 1.142 冪 0.00767 1 1 1 1 pq 0.546 0.454 n1 n2 70 60 p
^
^
冪 冢
冣
冪
冢
冣
Fail to reject H0.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
4. H0: p1 p2 ; Ha: p1 < p2 (claim) z0 1.645 共Lefttailed test) p
x1 x2 471 372 0.674 n1 n2 785 465
q 0.326
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.600 0.800兲 共0兲
^
冪p q冢n1 n1 冣 1
2
冪0.674 0.326冢7851 4651 冣
0.2 ⬇ 7.291 冪0.000752428
Reject H0. 5. H0: p1 p2 (claim); Ha: p1 > p2 z0 1.282 (Righttailed test) x1 x2 344 304 0.390 n1 n2 860 800 q 0.610 共 p p2兲 共 p1 p2兲 共0.400 0.380兲 共0兲 0.020 z 1 ⬇ 0.835 冪 0.0000574003 1 1 1 1 pq 0.390 0.610 n1 n2 860 800 p
^
^
冪 冢
冪
冣
冢
冣
Fail to reject H0. 6. H0: p1 p2 (claim); Ha: p1 p2 z0 ± 1.96 共Twotailed test兲 p
x1 x2 29 25 0.72 n1 n2 45 30
q 0.28
共 p1 p2兲 共 p1 p2兲 ^
z
共0.644 0.833兲 共0兲
^
冪p q冢n1 n1 冣 1
2
冪共0.72兲共0.28兲冢451 301 冣
0.189 ⬇ 1.786 冪0.0112
Fail to reject H0. 7. (a) H0: p1 p2 (claim); Ha: p1 p2 (b) z0 ± 1.96; Reject H0 if z < 1.96 or z > 1.96. p
x1 x2 520 865 0.385 n1 n2 1539 2055
q 0.615
共 p1 p2兲 共 p1 p2兲 ^
(c) z
^
冪p q冢n1 n1 冣 1
0.083 冪0.0000269069
2
共0.338 0.421兲 共0兲 1 1 冪0.385 0.615冢1539 2055 冣
⬇ 5.06
(d) Reject H0. (e) There is sufficient evidence at the 5% level to reject the claim that the proportion of adults using alternative medicines has not changed since 1991. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
235
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HYPOTHESIS TESTING WITH TWO SAMPLES
8. (a) H0: p1 p2 (claim); Ha: p1 p2 (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. p
x1 x2 5 19 0.149 n1 n2 77 84
q 0.851
共 p1 p2 兲 共 p1 p2 兲 ^
(c) z
^
冪p q冢n1 n1 冣 1
2
共0.06494 0.22620兲 共0兲
冪0.149 0.851冢771 841 冣
0.16126 ⬇ 2.870 冪0.0031563
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to reject the claim that the proportions of patients suffering new bouts of depression are the same for both groups. 9. (a) H0: p1 p2 (claim); Ha: p1 p2 (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. p
x1 x2 2201 2348 0.398 n1 n2 5240 6180
q 0.602
共 p1 p2兲 共 p1 p2兲 ^
(c) z
共0.4200 0.37994兲 0
^
冪p q冢n1 n1 冣 1
2
1 1 冪共0.398兲共0.602兲冢5240 6180 冣
0.041 4.362 冪0.000084494 (d) Reject H0. (e) There is sufficient evidence at the 10% significance level to reject the claim that the proportions of male and female senior citizens that eat the daily recommended number of servings of vegetables are the same. 10. (a) H0: p1 p2; Ha: p1 < p2 (claim) (b) z0 2.33; Reject H0 if z < 2.33. p
x1 x2 361 341 0.304 n1 n2 1245 1065
q 0.696
共 p1 p2 兲 共 p1 p2 兲 ^
(c) z
共0.2900 0.3202兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 共0.304兲共0.696兲 1245 1065
冣
0.0302 冪0.0003686
1.57
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that the proportion of male senior citizens eating the daily recommended number of servings of fruit is less than female senior citizens.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
11. (a) H0: p1 p2 ; Ha: p1 > p2 (claim) (b) z0 2.33; Reject H0 if z > 2.33. p
x1 x2 496 468 0.241 n1 n2 2000 2000
q 0.759
共 p1 p2兲 共 p1 p2兲 ^
(c) z
共0.248 0.234兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.241 0.759 2000 2000
冣
0.014 冪0.0001829
⬇ 1.04
(d) Fail to reject H0. (e) There is not sufficient evidence at the 1% significance level to support the claim that the proportion of adults who are smokers is greater in Alabama than in Missouri. 12. (a) H0: p1 p2 ; Ha: p1 < p2 (claim) (b) z0 1.645; Reject H0 if z < 1.645. p
x1 x2 228 185 0.1652 n1 n2 1500 1000
q 0.8348
共 p1 p2 兲 共 p1 p2 兲 ^
(c) z
共0.152 0.185兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.1652 0.8348 1500 1000
冣
0.033 冪0.0002298
⬇ 2.177
(d) Reject H0. (e) There is sufficient evidence at the 5% level to conclude that the proportion of adults who are smokers is lower in California than in Oregon. 13. (a) H0: p1 p2 ; Ha: p1 < p2 (claim) (b) z0 2.33; Reject H0 if z < 2.33. p
x1 x2 2083 985 0.216 n1 n2 9300 4900
q 0.784
共 p1 p2兲 共 p1 p2兲 ^
(c) z
^
冪p q 冢n1 n1 冣 1
2
共0.22398 0.20102兲 共0兲 1 1 冪0.216 0.784冢9300 4900 冣
0.02296 ⬇ 3.16 冪0.00005277
(d) Fail to reject H0. (e) There is sufficient evidence at the 1% significance level to support the claim that the proportion of twelfth grade males who said they had smoked in the last 30 days is less than the proportion of twelfth grade females. 14. (a) H0: p1 p2 (claim); Ha: p1 p2 (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. p
x1 x2 3444 3067 0.240 n1 n2 12,900 14,200
q 0.760 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
237
238
CHAPTER 8

共 p1 p2 兲 共 p1 p2 兲 ^
(c) z
HYPOTHESIS TESTING WITH TWO SAMPLES
冪p q冢n1 n1 冣 1
共0.267 0.216兲 共0兲
^
0.051 冪0.00002698
2
1 1 冪0.240 0.760冢12,900 14,200 冣
⬇ 9.818
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to reject the claim that the proportion of college students who said they had smoked in the last 30 days has not changed. 15. (a) H0: p1 p2 (claim); Ha: p1 p2 (b) z0 ± 1.96; Reject H0 if z < 1.96 or z > 1.96. p
x1 x2 805 746 0.705 n1 n2 1150 1050
q 0.295
共 p1 p2兲 共 p1 p2兲 ^
(c) z
共0.70000 0.71047兲 0
^
冪p q冢n1 n1 冣 1
2
1 1 冪共0.705兲共0.295兲冢1150 1050 冣
0.01047 0.54 冪0.0003789
(d) Fail to reject H0. (e) There is insufficient evidence at the 5% significance level to reject the claim that the proportions of Internet users are the same for both groups. 16. (a) H0: p1 p2; Ha: p1 > p2 (claim); (b) z0 1.645; Reject H0 if z > 1.645. (c) p
x1 x2 354 189 0.679 n1 n2 485 315
q 0.321
共 p1 p2 兲 共 p1 p2 兲 ^
z
^
冪p q冢n1 n1 冣 1
2
共0.730 0.600兲 0
冪共0.679兲共0.321兲冢4851 3151 冣
0.13 冪0.0011413
3.848 (d) Reject H0. (e) There is sufficient evidence at the 5% significance level to support the claim that the proportion of adults who use the Internet is greater for adults who live in an urban area than for adults who live in a rural area.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
17. H0: p1 p2 ; Ha: p1 < p2 (claim) z0 2.33 p
x1 x2 28 35 0.053 n1 n2 700 500
q 0.947
共 p1 p2兲 共 p1 p2兲 ^
z
共0.04 0.07兲 共0兲
^
冪p q冢n1 n1 冣 1
2
冪0.053 0.947冢7001 5001 冣
0.03 ⬇ 2.287 冪0.0001721
Fail to reject H0. There is insufficient evidence at the 1% significance level to support the claim. 18. H0: p1 p2 ; Ha: p1 > p2 (claim) z0 2.33 p
x1 x2 161 85 0.205 n1 n2 700 500
q 0.795
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.23 0.17兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.205 0.795 700 500
冣
0.06 冪0.0005588
⬇ 2.33
Reject H0. There is sufficient evidence at the 1% significance level to support the representatives belief. 19. H0: p1 p2; Ha: p1 < p2 (claim) z0 1.645 x x2 189 185 p 1 0.312 n1 n2 700 500 q 0.688
共 p1 p2兲 共 p1 p2兲 ^
z
共0.27 0.37兲 共0兲
^
冪p q冢n1 n1 冣 1
2
冪0.312 0.688冢7001 5001 冣
0.10 ⬇ 3.686 冪0.0007359
Reject H0. There is sufficient evidence at the 5% significance level to support the organization’s claim. 20. H0: p1 p2 (claim); Ha: p1 p2 z0 ± 1.96 p
x1 x2 224 145 0.308 n1 n2 700 500
q 0.692
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.32 0.29兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.308 0.692 700 500
冣
0.03 冪0.00073075
⬇ 1.110
Fail to reject H0. There is insufficient evidence at the 5% significance level to reject the claim.
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239
240
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
21. H0: p1 p2 ; Ha: p1 > p2 (claim) z0 1.645 p
x1 x2 7501 7501 0.548 n1 n2 13,300 14,100
q 0.452
共 p1 p2兲 共 p1 p2兲 ^
z
共0.564 0.532兲 共0兲
^
冪p q冢n1 n1 冣 1
2
1 1 冪共0.548兲共0.452兲冢13,300 14,100 冣
0.032 ⬇ 5.319 冪0.00003619
Reject H0. There is sufficient evidence at the 5% significance level to support the claim. 22. H0: p1 p2; Ha: p1 > p2 (claim) z0 1.645 p
x1 x2 5610 5934 0.428 n1 n2 13,200 13,800
q 0.572
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.425 0.430兲 共0兲
^
冪p q冢n1 n1 冣 1
2
1 1 冪共0.428兲共0.572兲冢13,200 13,800 冣
0.005 0.830 冪0.00003629
Fail to reject H0. There is not enough evidence at the 5% significance level to support the claim. 23. H0: p1 p2 (claim); Ha: p1 p2 z0 ± 2.576 p
x1 x2 7501 5610 0.495 n1 n2 13,300 13,200
q 0.505
共 p1 p2兲 共 p1 p2兲 ^
z
共0.564 0.425兲 共0兲
^
冪p q冢n1 n1 冣 1
2
1 1 冪共0.495兲共0.505兲冢13,300 13,200 冣
0.139 ⬇ 22,629 冪0.00003773
Reject H0. There is sufficient evidence at the 1% significance level to reject the claim. 24. H0: p1 p2 (claim); Ha: p1 p2 z0 ± 2.576 p
x1 x2 7501 5934 0.482 n1 n2 14,100 13,800
q 0.518
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.532 0.430兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 共0.482兲共0.518兲 14,100 13,800
冣
0.102 冪0.00003580
17.047
Reject H0. There is sufficient evidence at the 1% significance level to reject the claim.
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CHAPTER 8
冪pnq ^
25. 共 p1 p2 兲 ± zc ^
^
^
1 1 1
HYPOTHESIS TESTING WITH TWO SAMPLES
0.912 0.083 0.917 冪0.088 1,068,000 1,476,000
p2 q2 → 共0.088 0.083兲 ± 1.96 n2 ^

^
→ 0.005 ± 1.96 冪0.00000012671 → 0.004 < p1 p2 < 0.006
冪
26. 共 p1 p2 兲 ± zc ^
^
0.884 0.092 0.908 冪0.116 1,068,000 1,476,000
p1q1 p2 q2 → 共0.116 0.092兲 ± 1.96 n1 n2 ^
^
^
^
→ 0.024 ± 1.96冪0.0000001526 → 0.023 < p1 p2 < 0.025
CHAPTER 8 REVIEW EXERCISE SOLUTIONS 1. Independent because the two samples of laboratory mice are different. 2. Dependent, because the same mice were used for both experiments. 3. H0: 1 2 (claim); H1: 1 < 2 z0 1.645 z
共x1 x2 兲 共1 2 兲
冪
s 22
共1.28 1.34兲 共0兲
冪
0.06
⬇ 1.519 共0.30兲2 共0.23兲2 n1 n2 96 85 Fail to reject H0. There is not enough evidence to reject the claim. s 21
冪0.001560
4. H0: 1 2 (claim); Ha: 1 2 z0 ± 2.575 z
共x1 x2 兲 共 1 2 兲
冪n
s 21
1
s 22
n2
共5595 5575兲 共0兲 52兲 共68兲 冪共156 216 2
2
20 ⬇ 3.213 冪38.741
Reject H0. There is enough evidence to reject the claim. 5. H0: 1 2 ; H1: 1 < 2 (claim) z0 1.282 z
共x1 x2兲 共1 2兲
冪
共0.28 0.33兲 共0兲
冪
0.50 冪0.00058924
共0.11兲 共0.10兲 n1 n2 41 34 Reject H0. There is enough evidence to support the claim. s21
s22
2
2
⬇ 2.060
6. H0: 1 2; Ha: 1 2 (claim) z0 ± 1.96 z
共x1 x2 兲 共 1 2 兲
冪ns
2 1 1
s 22 n2
共87 85兲 共0兲 14兲 共15兲 冪共410 340 2
2
2 ⬇ 1.87 冪1.13981
Fail to reject H0. There is not enough evidence to support the claim.
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241
242

CHAPTER 8
HYPOTHESIS TESTING WITH TWO SAMPLES
7. (a) H0: 1 2 ; H1: 1 > 2 (claim) (b) z0 1.645; Reject H0 if z > 1.645. (c) z
共x1 x2兲 共1 2兲
冪n
s21
1
s22
共480 470兲 共0兲
冪
n2
共32兲 共54兲 36 41 2
2
10 冪99.566
⬇ 1.002
(d) Fail to reject H0. (e) There is not enough evidence to support the claim. 8. (a) H0: 1 2 ; Ha : 1 2 (claim) (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. (c) z
共x1 x2 兲 共 1 2 兲
冪n
s 21 1
s 22
n2
共360 380兲 共0兲
冪
共50兲2 共45兲2 38 35
20 冪123.647
⬇ 1.799
(d) Reject H0. (e) There is sufficient evidence at the 10% significance level to support the claim that the caloric content of the two types of french fries is different. 9. H0: 1 2 (claim); Ha: 1 2 d.f. n1 n2 2 29 t0 ± 2.045 t
共x1 x2兲 共1 2兲
冪共n
1
1兲 共n2 1兲 n1 n2 2 s 21
s22
共300 290兲 共0兲
冪n1 n1 冪共19 1兲共1926兲 12共12 2 1兲共22兲 冪191 121 1
2
2
2
10 ⬇ 1.104 冪603.172 冪0.1360 Fail to reject H0. There is not enough evidence to reject the claim. 10. H0: 1 2 (claim); Ha: 1 2 d.f. min{n1 1, n2 1} 5 t0 ± 2.015 t
共x1 x2 兲 共 1 2兲
冪n
s 12
1
s22 n2
共0.015 0.019兲 共0兲
冪
共0.011兲 共0.004兲 8 6 2
2
.004 冪0.0000178
⬇ 0.948
Fail to reject H0. There is not enough evidence to reject the claim.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
11. H0: 1 2 (claim); Ha: 1 > 2 d.f. min再n1 1, n2 1冎 24 t0 1.711
共x1 x2兲 共1 2 兲
冪
s 21
s22
共183.5 184.7兲 共0兲
冪
2
2
1.2
⬇ 1.460 共1.3兲 共3.9兲 n1 n2 25 25 Fail to reject H0. There is not enough evidence to reject the claim. t
冪0.676
12. H0: 1 2 (claim); Ha: 1 < 2 d.f. n1 n2 2 39 t0 2.326 t
共x1 x2 兲 共 1 2 兲
冪
共n1 1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共24.5 26.4 兲 共0 兲
冪
冪
1.9 冪6.607 冪0.1026
共19 1兲共2.95兲 共20 1兲共2.15兲 1 1 19 20 2 19 20 Fail to reject H0. There is not enough evidence to reject the claim. 2
2
⬇ 2.308
13. H0: 1 2 ; Ha: 1 2 (claim) d.f. n1 n2 2 10 t0 ± 3.169 t
共x1 x2兲 共1 2兲
冪共n
1
1兲 s 21 共n2 1兲s22 n1 n2 2
6 冪5.22 冪0.343
冪
1 1 n1 n2
共61 55兲 共0兲 共7 1兲 1.2 冪共5 1兲 3.3 572 冪15 17 2
2
⬇ 4.484
Reject H0. There is enough evidence to support the claim. 14. H0: 1 2 (claim); Ha: 1 < 2 d.f. min{n1 1, n2 1} 5 t0 1.476 t
共x1 x2 兲 共 1 2 兲
冪ns
2 1 1
s 22 n2
共520 500兲 共0兲
冪共257 兲
2
共55兲2 6
20 ⬇ 0.821 冪593.452
Fail to reject H0. There is not enough evidence to reject the claim. 15. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) d.f. n1 n2 2 42 t0 1.645; Reject H0 if t > 1.645.
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243
244
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
(c) x1 51.476, s1 11.007, n1 21 x2 41.522, s2 17.149, n2 23 共x1 x2兲 共1 2兲 t 共n 1 1兲s 21 共n2 1兲s22 n1 n2 2
冪
9.954 冪211.7386 冪0.0911
共51.476 41.522兲 共0兲 兲 共23 1兲共17.149兲 冪共21 1兲共11.007 冪211 231 21 23 2 2
2
⬇ 2.266
(d) Reject H0. (e) There is sufficient evidence at the 5% significance level to support the claim that the third graders taught with the directed reading activities scored higher than those taught without the activities. 16. (a) H0: 1 2 (claim); Ha: 1 2 (b) d.f. n1 n2 2 20 t0 ± 2.086; Reject H0 if t < 2.086 or t > 2.086. (c) t
共x1 x2 兲 共 1 2 兲
冪
共n1 1兲s21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
共32,750 31,200兲 共0兲 兲 共10 1兲共1825兲 冪共12 1兲共1900 冪121 101 12 10 2 2
2
750 ⬇ 1.939 冪3,484,281.25 冪0.183
(d) Fail to reject H0. There is not enough evidence to reject the claim. 17. H0: d 0 (claim); Ha: d 0
0.05 and d.f. n 1 99 t0 ± 1.96 (Twotailed test) t
d d 10 0 10 ⬇ 8.065 sd 12.4 1.24 冪n 冪100
Reject H0. 18. H0: d 0; Ha: d < 0 (claim)
0.01 and d.f. n 1 24 t0 2.492 (Lefttailed test) t
d d 3.2 0 3.2 ⬇ 11.594 sd 1.38 0.276 冪n 冪25
Fail to reject H0.
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
19. H0: d 6 (claim); Ha: d > 6
0.10 and d.f. n 1 32 t0 1.282 (Righttailed test) t
4.3 d d 10.3 6 ⬇ 19.921 sd 1.24 0.21586 冪n 冪33
Reject H0. 20. H0: d 0; Ha: d 15 (claim)
0.05 and d.f. n 1 36 t0 ± 1.96 (Twotailed test) t
d d 17.5 15 2.5 ⬇ 3.755 sd 4.05 0.6691 冪n 冪37
Reject H0. 21. (a) H0: d 0; Ha: d > 0 (claim) (b) t0 1.383; Reject H0 if t > 1.383. (c) d 5 and sd ⬇ 8.743 (d) t
d d 5 0 5 ⬇ 1.808 sd 8.743 2.765 冪n 冪10
(e) Reject H0. (f) There is enough evidence to support the claim. 22. (a) H0: d 0; Ha: d > 0 (claim) (b) t0 1.372; Reject H0 if t > 1.372. (c) d ⬇ 0.636 and sd ⬇ 5.870 (d) t
d d 0.636 0 0.636 ⬇ 0.359 sd 5.870 1.7699 冪n 冪11
(e) Fail to reject H0. (f) There is not enough evidence to support the claim.
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245
246
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
23. H0: p1 p2 ; Ha: p1 p2 (claim) z0 ± 1.96 (Twotailed test) p
x1 x2 425 410 0.522 n1 n2 840 760
q 0.478
共 p1 p2兲 共 p1 p2兲 ^
z
共0.506 0.539兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.522 0.478 840 760
冣
0.033 冪0.0006254
⬇ 1.320
Fail to reject H0. 24. H0: p1 p2 (claim); Ha: p1 > p2 z0 2.33 (Righttailed test) p
x1 x2 36 46 0.273 n1 n2 100 200
q 0.727
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.360 0.230兲 共0兲
^
冪p q冢n1 n1 冣 1
2
冪0.273 0.727冢1001 2001 冣
0.130 ⬇ 2.383 冪0.002977
0.04 ⬇ 1.293 冪0.0009576
Reject H0. 25. H0: p1 p2 ; Ha: p1 > p2 (claim) z0 1.282 (Righttailed test) p
x1 x2 261 207 0.450 n1 n2 556 483
q 0.550 共 p p2兲 共 p1 p2兲 z 1 1 1 pq n1 n2 ^
共0.469 0.429兲 共0兲
^
冪 冢
冣
冪0.450 0.550冢5561 4831 冣
Reject H0. 26. H0: p1 p2 ; Ha: p1 < p2 (claim) z0 1.645 (Lefttailed test) p
x1 x2 86 107 ⬇ 0.092 n1 n2 900 1200
q 0.908
共 p1 p2 兲 共 p1 p2 兲 ^
z
共0.096 0.089兲 共0兲
^
冪 冢
1 1 pq n1 n2
冣
冪
冢
1 1 0.092 0.908 900 1200
冣
0.007 冪0.00016243
⬇ 0.549
Fail to reject H0.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
27. (a) H0: p1 p2 (claim); Ha: p1 p2 (b) z0 ± 1.645; Reject H0 if z < 1.645 or z > 1.645. p
x1 x2 398 530 0.516 n1 n2 800 1000
q 0.484
共 p1 p2兲 共 p1 p2兲 ^
(c) z
共0.4975 0.5300兲 共0兲
^
冪p q冢n1 n1 冣 1
2
1 冪0.516 0.484冢8001 1000 冣
0.0325 ⬇ 1.371 冪0.0005619
(d) Fail to reject H0. (e) There is not enough evidence to reject the claim. 28. (a) H0: p1 p2 ; Ha: p1 > p2 (claim) (b) z0 1.645; Reject H0 if z > 1.645. p
x1 x2 986 5576 ⬇ 0.134 n1 n2 6164 42,890
q 0.866
共 p1 p2 兲 共 p1 p2 兲 ^
(c) z
冪p q 冢n1 n1 冣 1
共0.160 0.130兲 共0兲
^
0.030 冪0.00002153
2
1 1 冪0.134 0.866冢6164 42,890 冣
⬇ 6.465
(d) Reject H0. (e) There is enough evidence to support the claim.
CHAPTER 8 QUIZ SOLUTIONS 1. (a) H0: 1 2 ; Ha: 1 > 2 (claim) (b) n1 and n2 > 30 and the samples are independent → Right tailed ztest (c) z0 1.645; Reject H0 if z > 1.645. (d) z
共x1 x2兲 共1 2兲
冪ns
2 1
1
s22 n2
共149 145兲 共0兲
冪共3549兲
2
共33兲2 50
4 ⬇ 0.585 冪46.780
(e) Fail to reject H0. (f) There is not enough evidence at the 5% significance level to support the claim that the mean score on the science assessment for male high school students was higher than for the female high school students. 2. (a) H0: 1 2 (claim); Ha: 1 2 (b) n1 and n2 < 30, the samples are independent, and the populations are normally distributed. → Twotailed ttest (assume variances are equal)
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247
248
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
(c) d.f. n1 n2 2 26 t0 ± 2.779; Reject H0 if t < 2.779 or t > 2.779.
共x1 x2兲 共 1 2 兲
(d) t
冪共n
1
1兲 共n2 1兲 n1 n2 2 s 21
s22
共153 149兲 共0兲
冪共13 1兲共1332兲 15共15 2 1兲共30兲 冪131 151 2
2
4.0 ⬇ 0.341 冪957.23 冪0.14359
(e) Fail to reject H0. (f) There is not enough evidence at the 1% significance level to reject the teacher’s claim that the mean scores on the science assessment test are the same for fourth grade boys and girls. 3. (a) H0: p1 p2 ; Ha: p1 > p2 (claim) (b) Testing 2 proportions, n1 p, n1 q, n2 p, and n2 q 5, and the samples are independent → Righttailed ztest (c) z0 1.28; Reject H0 if z > 1.28. (d) p
x1 x2 2043 3018 0.288 n1 n2 6382 11,179
q 0.712
共 p1 p2兲 共p1 p2兲 ^
z
共0.32 0.27兲 共0兲
^
冪p q冢n1 n1 冣 1
2
1 1 冪0.288 0.712冢6382 11,179 冣
0.05 ⬇ 7.04 冪0.000050473 (e) Reject H0. (f) There is sufficient evidence at the 10% significance level to support the claim that the proportion of fatal crashes involving alcohol is higher for drivers in the 21 to 24 age group than for drivers ages 25 to 35. 4. (a) H0: d 0; Ha: d < 0 (claim) (b) Dependent samples and both populations are normally distributed. → onetailed ttest (c) t0 1.796; Reject H0 if t < 1.796. (d) t
d d 0 68.5 68.5 ⬇ 9.016 sd 26.318 7.597 冪n 冪12
(e) Reject H0. (f) There is sufficient evidence at the 5% significance level to conclude that the students’ SAT scores improved on the second test.
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CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
CUMULATIVE REVIEW, CHAPTERS 6–8 1. (a) p ^
x 570 0.570 n 1000
q 0.430 ^
np 570 5 ^
nq 430 5 ^
Use normal distribution pq p ± zc 0.570 ± 1.96 n
冪
^
兲共0.430兲 0.570 ± 0.031 ⇒ 共0.539, 0.601兲 冪共0.5701000
^ ^
(b) Given that the 95% CI is 共0.539, 0.601兲, it is unlikely that more than 60% of adults believe it is somewhat or very likely that life exists on other planets. 2. H0: d 0 Ha: d > 0 (claim) d 6.833 sd 3.713 d.f. n 1 11 t 0 1.796 t
d 0 6.833 0 6.375 sd 3.713 冪n 冪12
Reject H0. There is enough evidence to support the claim. 3. x ± zc 4. x ± t c
s 冪n
s 冪n
5. x ± z c 6. x ± t c
s 冪n
s 冪n
3.4
29.97 ± 1.96 3.46 ± 1.753
1.63
3.46 ± 0.71 ⇒ 共2.75, 4.17兲; tdistribution
冪16
12.1 ± 2.787 8.21 ± 2.365
26.97 ± 1.03 ⇒ 共25.94, 28.00兲; zdistribution
冪42
2.64 冪26
12.1 ± 1.4 ⇒ 共10.7, 13.5兲; tdistribution
0.62 8.21 ± 0.52 ⇒ 共7.69, 8.73兲; tdistribution 冪8
7. H0: 1 2 Ha: 1 > 2 (claim) z 1.282 共x x2兲 0 z 1 s 21 s 22 n1 n2
冪
3086 2263
冪
共563兲 共624兲 85 68 2
2
823 冪9790.282
8.318
Reject H0. There is enough evidence to support the claim.
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249
250

CHAPTER 8
HYPOTHESIS TESTING WITH TWO SAMPLES
8. H0: 33
9. H0: p 0.19 (claim)
Ha: < 33 (claim)
Ha: p < 0.19
10. H0: 0.63 (claim)
11. H0: 2.28
Ha: 0.63 12. (a)
(b)
Ha: 2.28 (claim)
共n 1兲s2 共n 1兲s2 共26 1兲共3.1兲2 共26 1兲共3.1兲2 < 2 < ⇒ < 2 < 2 2 R L 46.928 10.520 ⇒ 5.1 < < 22.8
共n 1兲 R2
s2
共n 1兲 ⇒ 冪5.1 < < 冪22.8 L2 ⇒ 2.3 < < 4.8 s2
< <
(c) Because the 99% CI of is 共2.3, 4.8兲 and contains 2.5, there is not enough evidence to support the pharmacist’s claim. 13. H0: 1 2 Ha: 1 < 2 (claim) d.f. n1 n2 2 15 15 2 28 t0 1.701 t
共x1 x2兲 0
冪共n
1
1兲s 21 共n2 1兲s22 n1 n2 2
冪n1 n1 1
2
57.9 61.1
冪共15 1兲共150.8兲 15共15 2 1兲共0.6兲 冪151 151 2
3.2 冪0.5冪0.133
2
12.394
Reject H0. There is enough evidence to support the claim. 14. (a) x 296.231 s 111.533 x tc
s 冪n
⇒ 296.231 ± 2.060
111.533 冪26
⇒ 296.231 ± 45.059 ⇒ 共251.2, 341.3兲 (b) H0: 280 Ha: < 280 (claim) t0 1.316 t
x 280 296.231 280 0.742 s 111.533 冪n 冪26
Fail to reject H0. There is not enough evidence to support the claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 8

HYPOTHESIS TESTING WITH TWO SAMPLES
15. (a) H0: p1 p 2 (claim) Ha: p1 p 2 p
x1 x2 195 204 0.621 n1 n2 319 323
q 0.379 z0 ± 1.645
共 p1 p2兲 0 ^
z
^
冪p q冢n1 n1 冣 1
2
共0.611 0.632兲
冪共0.621兲共0.379兲冢3191 3231 冣 0.021 冪0.001466
0.548
Fail to reject H0. There is not enough evidence to reject the claim. 16. x ± z
s 冪n
⇒ 150 ± 1.645
34 冪120
⇒ 150 ± 5 ⇒ 共145, 155兲
(b) H0: 145 (claim) Ha: < 145 z0 1.645 z
x 145 150 145 1.611 s 34 冪n 冪120
Fail to reject H0. There is not enough evidence to reject the claim.
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251
CHAPTER
Correlation and Regression
9
9.1 CORRELATION
9.1 Try It Yourself Solutions 1ab.
c. Yes, it appears that there is a negative linear correlation. As family income increases, the percent of income donated to charity decreases.
Donating percent
y 10 8 6 4 2 x 40 45 50 55 60 65 70 75
Family income (in thousands of dollars)
y
2ab.
c. No, it appears that there is no linear correlation between age and subscriptions.
6
Subscriptions
5 4 3 2 1 x 10 20 30 40 50 60 70
Age (in years)
3ab.
2000
c. Yes, there appears to be a positive linear relationship between budget and worldwide gross.
0
225 0
4a. n ⫽ 6 b.
x
y
xy
x2
y2
42 48 50 59 65 72
9 10 8 5 6 3
378 480 400 295 390 216
1764 2304 2500 3481 4225 5184
81 100 64 25 36 9
兺x ⫽ 336
兺y ⫽ 41
兺xy ⫽ 2159
兺x 2 ⫽ 19,458
兺y2 ⫽ 315
c. r ⫽ ⫽
n兺xy ⫺ 共兺x兲共兺y兲 冪n兺x2 ⫺ 共兺x兲2冪n兺y2 ⫺ 共兺y兲2
⫽
6共2159兲 ⫺ 共336兲共41兲 冪6共19,458兲 ⫺ 共336兲2冪6共315兲 ⫺ 共41兲2
⫺822 ⬇ ⫺0.916 冪3852 冪209
d. Because r is close to ⫺1, there appears to be a strong negative linear correlation between income level and donating percent.
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254
CHAPTER 9

CORRELATION AND REGRESSION
5a. Enter the data. b. r ⬇ 0.838 c. Because r is close to 1, there appears to be a strong positive linear correlation between budgets and worldwide grosses. 6a. n ⫽ 6 b. ␣ ⫽ 0.01 c. 0.917
ⱍⱍ
d. Because r ⬇ 0.916 < 0.917, the correlation is not significant. e. There is not enough evidence to conclude that there is a significant linear correlation between income level and the donating percent. 7a. H0 : ⫽ 0; Ha : ⫽ 0 b. ␣ ⫽ 0.01 c. d.f. ⫽ n ⫺ 2 ⫽ 23 d. ± 2.807; Reject H0 if t < ⫺2.807 or t > 2.807. e. t ⫽
r
冪
1⫺ n⫺2
r2
⫽
0.83773
冪
1 ⫺ 共0.83773兲 25 ⫺ 2
2
⫽
0.83773 冪0.012966
⬇ 7.357
f. Reject H0. g. There is enough evidence in the sample to conclude that a significant linear correlation exists.
9.1 EXERCISE SOLUTIONS 1. r ⫽ ⫺0.925 represents a stronger correlation because it is closer to ⫺1 than r ⫽ 0.834 is to ⫹1. 2. (c) Correlation coefficients must be contained in the interval 关⫺1, 1兴. 3. A table can be used to compare r with a critical value or a hypotheses test can be performed using a ttest. 4. r ⫽ sample correlation coefficient ⫽ population correlation coefficient 5. Negative linear correlation
6. No linear correlation
7. No linear correlation
8. Positive linear correlation
9. (c), You would expect a positive linear correlation between age and income. 10. (d), You would not expect age and height to be correlated. 11. (b), You would expect a negative linear correlation between age and balance on student loans. 12. (a), You would expect the relationship between age and body temperature to be fairly constant.
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CHAPTER 9

CORRELATION AND REGRESSION
13. Explanatory variable: Amount of water consumed Response variable: Weight loss 14. Explanatory variable: Hours of safety classes Response variable: Number of accidents y
Systolic blood pressure
15. (a) 200 150 100 50
x 10 20 30 40 50 60 70
Age (in years)
(b)
x
y
xy
x2
y2
16 25 39 45 49 64 70 29 57 20
109 122 143 132 199 185 199 130 175 118
1744 3050 5577 5940 9751 11,840 13,930 3770 9975 2360
256 625 1521 2025 2401 4096 4900 841 3249 400
11,881 14,884 20,449 17,424 39,601 34,225 39,601 16,900 30,625 13,924
兺x ⫽ 414
兺y ⫽ 1512
兺xy ⫽ 67,937
兺x2 ⫽ 20,314
兺y2 ⫽ 239,514
r⫽ ⫽
n兺xy ⫺ 共兺x兲共兺y兲 冪n兺x2 ⫺ 共兺x兲2冪n兺y2 ⫺ 共兺y兲2
53,402
⫽
10共67,937兲 ⫺ 共414兲共1512兲 冪10共20,314兲 ⫺ 共414兲2冪10共239,514兲 ⫺ 共1512兲2
⬇ 0.908
冪31,744 冪108,996
(c) Strong positive linear correlation y
16. (a) Vocabulary size
2500 2000 1500 1000 500 x 1
2
3
4
5
6
Age (in years)
(b)
x
y
xy
x2
y2
1 2 3 4 5 6 3 5 2 4 6
3 400 1200 1500 2100 2600 1100 2000 500 1525 2500
3 880 3600 6000 10,500 15,600 3300 10,000 1000 6100 15,000
1 4 9 16 25 36 9 25 4 16 36
9 193,600 1,440,000 2,250,000 4,410,000 6,760,000 1,210,000 4,000,000 250,000 2,325,625 6,250,000
兺x ⫽ 41
兺y ⫽ 15,468
兺xy ⫽ 71,983
兺x 2 ⫽ 181
兺y 2 ⫽ 29,089,234
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255
256
CHAPTER 9
r⫽ ⫽

CORRELATION AND REGRESSION
n兺xy共兺x兲共兺y兲
冪
n兺x2
2
⫺ 共兺x兲
冪n兺y 2 ⫺ 共兺y兲2
11共71,983兲 ⫺ 共41兲共15,468兲 冪11共181兲 ⫺ 共41兲2 冪11共29,089,234兲 ⫺ 共15,468兲2
⫽
157,625 冪310 冪80,722,550
⬇ 0.996
(c) Strong positive linear correlation y
17. (a) Test score
100 80 60 40 20 x 2
4
6
8
Hours studying
(b)
x
y
xy
x2
y2
0 1 2 4 4 5 5 5 6 6 7 7 8
40 41 51 48 64 69 73 75 68 93 84 90 95
0 41 102 192 256 345 365 375 408 558 588 630 760
0 1 4 16 16 25 25 25 36 36 49 49 64
1600 1681 2601 2304 4096 4761 5329 5625 4624 8649 7056 8100 9025
兺x ⫽ 60
兺y ⫽ 891
兺xy ⫽ 4620
兺x2 ⫽ 346
兺y2 ⫽ 65,451
r⫽ ⫽
n兺xy ⫺ 共兺x兲共兺y兲 13共4620兲 ⫺ 共60兲共891兲 ⫽ 2 2 2 冪13共346兲 ⫺ 共60兲2冪13共65,451兲 ⫺ 共891兲2 ⫺ 共兺x兲 冪n兺y ⫺ 共兺y兲
冪n兺x2
6600 冪898冪56,982
⬇ 0.923
(c) Strong positive linear correlation y
18. (a) Test score
100 80 60 40 20 x 2
4
6
8
10
Hours spent online
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CHAPTER 9
(b)
x
y
xy
x2
y2
0 1 2 3 3 5 5 5 6 7 7 10
96 85 82 74 95 68 76 84 58 65 75 50
0 85 164 222 285 340 380 420 348 455 525 500
0 1 4 9 9 25 25 25 36 49 49 100
9216 7225 6724 5476 9025 4624 5776 7056 3364 4225 5625 2500
兺x ⫽ 54
兺y ⫽ 908
兺xy ⫽ 3724
兺x 2 ⫽ 332
兺y 2 ⫽ 70,836
r⫽ ⫽
冪n兺x2

CORRELATION AND REGRESSION
n兺xy ⫺ 共 兺x兲共兺y兲 ⫺ 共兺x兲2 冪n兺y 2 ⫺ 共兺y兲2 12共3724兲 ⫺ 共54兲共908兲
冪12共332兲 ⫺ 共54兲2 冪12共70836兲 ⫺ 共908兲2
⫽
⫺4344 冪1068 冪25,568
⬇ ⫺0.831
(c) Strong negative linear correlation y
Gross (in millions of dollars)
19. (a)
2000 1500 1000 500 x 170
190
210
Budget (in millions of dollars)
(b)
x
y
xy
x2
y2
207 204 200 200 180 175 175 170
553 391 1835 784 749 218 255 433
114,471 79,764 367,000 156,800 134,820 38,150 44,625 73,610
42,849 41,616 40,000 40,000 32,400 30,625 30,625 28,900
305,809 152,881 3.37 ⫻ 106 614,656 561,001 47,524 65,025 187,489
兺x ⫽ 1511
兺y ⫽ 5218
兺xy ⫽ 1,009,240
兺x2 ⫽ 287,015
兺y2 ⫽ 5,301,610
r⫽ ⫽
n兺xy ⫺ 共兺x兲共兺y兲 冪n兺x2 ⫺ 共兺x兲2 冪n兺y 2 ⫺ 共兺y兲2
⫽
8共1,009,240兲 ⫺ 共1511兲共5218兲 冪8共287,015兲 ⫺ 共1511兲2 冪8共5,301,610兲 ⫺ 共5218兲2
189,522 ⬇ 0.427 冪12,999 冪15,185,356
(c) Weak positive linear correlation
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257
258
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CORRELATION AND REGRESSION
y
Speed of sound (in feet per second)
20. (a) 1150 1100 1050 1000 950
x 10 20 30 40 50
Altitude (in thousands of feet)
(b)
x
y
0 5 10 15 20 25 30 35 40 45 50
1116.3 1096.9 1077.3 1057.2 1036.8 1015.8 994.5 969.0 967.7 967.7 967.7
兺x ⫽ 275
兺y ⫽ 11,266.9
r⫽ ⫽
x2
xy 0.000 5484.5 10,773 15,858 20,736 25,395 29,835 33,915 38,708 43,547 48,385
0 25 100 225 400 625 900 1225 1600 2025 2500
兺xy ⫽ 272,636
n兺xy ⫺ 共兺x兲共兺y兲 冪n兺x2 ⫺ 共兺x兲2冪n兺y2 ⫺ 共兺y兲2
⫽
兺x2 ⫽ 9625
y2 1.25 ⫻ 106 1.2 ⫻ 106 1.16 ⫻ 106 1.12 ⫻ 106 1.07 ⫻ 106 1.03 ⫻ 106 989,030 938,961 936,443 936,443 936,443 兺y2 ⫽ 11,571,687.43
11共272,636兲 ⫺ 共275兲共11,266.9兲 冪11共9625兲 ⫺ 共275兲2冪11共11,571,687.43兲 ⫺ 共11,266.9兲2
⫺99,401.5 ⬇ ⫺0.972 冪30,250冪345,526.12
(c) Strong negative linear correlation y
Dividends per share
21. (a) 2.0 1.5 1.0 0.5
x 1
2
3
4
5
6
Earnings per share
(b)
x2
y2
x
y
xy
2.34 1.96 1.39 3.07 0.65 5.21 0.88 3.23 2.54 1.03
1.33 1.07 1.15 0.25 1.00 1.00 1.59 1.20 1.62 0.20
3.112 2.097 1.599 0.768 0.650 5.210 1.399 3.876 4.115 0.206
5.476 3.842 1.932 9.425 0.423 27.144 0.774 10.433 6.452 1.061
1.769 1.145 1.323 0.063 1.000 1.000 2.528 1.440 2.624 0.040
兺x ⫽ 22.3
兺y ⫽ 10.41
兺xy ⫽ 23.031
兺x2 ⫽ 66.961
兺y2 ⫽ 12.931
r⫽ ⫽
n兺xy ⫺ 共兺x兲共兺y兲 10共23.031兲 ⫺ 共22.3兲共10.41兲 ⫽ 2 2 2 冪10共66.961兲 ⫺ 共22.3兲2冪10共12.931兲 ⫺ 共10.41兲2 ⫺ 共兺x兲 冪n兺y ⫺ 共兺y兲
冪n兺x2
⫺1.833 ⫽ ⫺0.031 冪172.32冪20.942
(c) No linear correlation © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 9

CORRELATION AND REGRESSION
y
Arrests (in millions)
22. (a) 0.80 0.75 0.70 0.65 0.60 0.55
x 1.1
1.3
1.5
1.7
Crimes (in millions)
(b)
x
y
xy
x2
y2
1.66 1.65 1.60 1.55 1.44 1.40 1.32 1.23 1.22 1.23 1.22 1.18 1.16 1.19
0.72 0.72 0.78 0.80 0.73 0.72 0.68 0.64 0.63 0.63 0.62 0.60 0.59 0.60
1.195 1.188 1.248 1.240 1.051 1.008 0.898 0.787 0.769 0.775 0.756 0.708 0.684 0.714
2.756 2.723 2.560 2.403 2.074 1.960 1.742 1.513 1.488 1.513 1.488 1.392 1.346 1.416
0.518 0.518 0.608 0.640 0.533 0.518 0.462 0.410 0.397 0.397 0.384 0.360 0.348 0.360
兺x ⫽ 19.05
兺y ⫽ 9.46
兺xy ⫽ 13.021
兺x 2 ⫽ 26.373
兺y 2 ⫽ 6.455
r⫽ ⫽
冪n兺x2
n兺xy ⫺ 共兺x兲共兺y兲 ⫺ 共兺x兲2 冪n兺y 2 ⫺ 共兺y兲2 14共13.021兲 ⫺ 共19.05兲共9.46兲
冪14共26.373兲 ⫺ 共19.05兲2 冪14共6.455兲 ⫺ 共9.46兲2
⫽
2.081 冪6.320 冪0.878
⫽ 0.883 (c) Strong positive linear correlation 23. r ⬇ 0.623 n ⫽ 8 and ␣ ⫽ 0.01 cv ⫽ 0.834 r ⬇ 0.623 < 0.834 ⇒ The correlation is not significant.
ⱍⱍ
or H0: ⫽ 0; and Ha: ⫽ 0 ␣ ⫽ 0.01 d.f. ⫽ n ⫺ 2 ⫽ 6 cu ⫽ ± 3.703; Reject H0 if t < ⫺3.707 or t > 3.707. t⫽
r
冪
1⫺ n⫺2
r2
⫽
0.623
冪
1 ⫺ 共0.623兲 8⫺2
2
⫽
0.623 冪0.10198
⫽ 1.951
Fail to reject H0 . There is not enough evidence to conclude that there is a significant linear correlation between vehicle weight and the variability in braking distance.
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259
260

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CORRELATION AND REGRESSION
24. r ⬇ 0.955 n ⫽ 8 and ␣ ⫽ 0.05 cv ⫽ 0.707
ⱍrⱍ ⫽ 0.955 > 0.707 ⇒ The correlation is significant. or H0 : p ⫽ 0; Ha: p ⫽ 0
␣ ⫽ 0.05 d.f. ⫽ n ⫺ 2 ⫽ 6 cu ⫽ ± 2.447; Reject H0 if t < ⫺2.447 or t > 2.447. t⫽
r
冪
1⫺r n⫺2
2
⫽
0.955
冪
1 ⫺ 共0.955兲 8⫺2
2
⫽
0.955 冪0.01466
⫽ 7.887
Reject H0. There is enough evidence to conclude that there is a significant linear correlation between vehicle weight and variability in braking distance. 25. r ⬇ 0.923 n ⫽ 8 and ␣ ⫽ 0.01 cv ⫽ 0.834 r ⬇ 0.923 > 0.834 ⇒ The correlation is significant. or
ⱍⱍ
H0 : ⫽ 0; Ha: ⫽ 0 ␣ ⫽ 0.01 d.f. ⫽ n ⫺ 2 ⫽ 11 cu ⫽ ± 3.106; Reject H0 if t < ⫺3.106 or t > 3.106. r ⬇ 0.923 t⫽
r
0.923
⫽
兲 冪1n ⫺⫺ 2 冪1 ⫺13共0.923 ⫺2 r2
2
⫽
0.923 ⬇ 7.955 冪0.01346
Reject H0 . There is enough evidence to conclude that a significant linear correlation exists. 26. r ⬇ 0.831 n ⫽ 12 and ␣ ⫽ 0.05 cv ⫽ 0.576 r ⫽ 0.831 > 0.576 ⇒ The correlation is significant. or H0: ⫽ 0; Ha: ⫽ 0 ␣ ⫽ 0.05 d.f. ⫽ n ⫺ 2 ⫽ 10 cu ⫽ ± 2.228; Reject H0 if t < ⫺2.228 or t > 2.228. r ⬇ ⫺0.831 r ⫺0.831 t⫽ ⫽ ⬇ ⫺4.724 1 ⫺ r2 1 ⫺ 共⫺0.831兲2 n⫺2 12 ⫺ 2
ⱍⱍ
冪
冪
Reject H0. There is enough evidence to conclude that there is a significant linear correlation between the data. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 9

CORRELATION AND REGRESSION
27. r ⬇ ⫺0.030 n ⫽ 10 and ␣ ⫽ 0.01 cv ⫽ 0.765 r ⬇ 0.030 < 0.765 ⇒ The correlation is not significant.
ⱍⱍ
or H0 : ⫽ 0; Ha: ⫽ 0 ␣ ⫽ 0.01 d.f. ⫽ n ⫺ 2 ⫽ 8 cu ⫽ ± 3.355; Reject H0 if t < ⫺3.355 or t > 3.355. r ⬇ ⫺0.030 r t⫽ ⫽ 1 ⫺ r2 n⫺2
冪
⫽
⫺0.030
兲 冪1 ⫺10共⫺0.030 ⫺2
2
⫺0.030 ⫽ ⫺0.085 冪0.125
Fail to reject H0 . There is not enough evidence at the 1% significance level to conclude there is a significant linear correlation between earnings per share and dividends per share. 28. r ⬇ 0.842 n ⫽ 11 and ␣ ⫽ 0.05 cv ⫽ 0.602 r ⫽ 0.842 > 0.602 ⇒ The correlation is significant.
ⱍⱍ
or H0: ⫽ 0; Ha: ⫽ 0 ␣ ⫽ 0.05 d.f. ⫽ n ⫺ 2 ⫽ 9 cu ⫽ ± 2.262; Reject H0 if t < ⫺2.262 or t > 2.262. r ⬇ 0.842 r 0.832 t⫽ ⫽ ⫽ 4.682 2 1⫺r 1 ⫺ 共⫺0.842兲2 n⫺2 11 ⫺ 2
冪
冪
Reject H0. There is enough evidence to conclude that there is a significant correlation between the data. 29. The correlation coefficient remains unchanged when the xvalues and yvalues are switched. 30. The correlation coefficient remains unchanged when the xvalues and yvalues are switched. 31. Answers will vary.
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261
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CHAPTER 9

CORRELATION AND REGRESSION
9.2 LINEAR REGRESSION
9.2 Try It Yourself Solutions 1a. n ⫽ 6 x
y
xy
x2
42 48 50 59 65 72
9 10 8 5 6 3
378 480 400 295 390 216
1764 2304 2500 3481 4225 5184
兺x ⫽ 336
兺y ⫽ 41
兺xy ⫽ 2159
兺x 2 ⫽ 19,458
b. m ⫽
n兺xy ⫺ 共兺x兲共兺y兲 6共2159兲 ⫺ 共336兲共41兲 ⫺822 ⫽ ⫽ ⬇ ⫺0.2133956 n兺x2 ⫺ 共兺x兲2 6共19,458兲 ⫺ 共336兲2 3852 ⬇ 18.7837 冢416冣 ⫺ 共⫺0.2133956兲冢336 6 冣
c. b ⫽ y ⫺ mx ⫽
d. y ⫽ ⫺0.213x ⫹ 18.783 ^
2a. Enter the data. b. m ⬇ 10.93477; b ⬇ ⫺710.61551 c. y ⫽ 10.935x ⫺ 710.616 ^
3a. (1) y ⫽ 12.481共2兲 ⫹ 33.683
(2) y ⫽ 12.481共3.32兲 ⫹ 33.683
^
b. (1) 58.645
^
(2) 75.120
c. (1) 58.645 minutes
(2) 75.120 minutes
9.2 EXERCISE SOLUTIONS 1. c
2. a
3. d
4. b
5. g
6. e
7. h
8. f
9. c
10. b
11. a
12. d
y
xy
764 625 520 510 492 484 450 430 410
55 47 51 28 39 34 33 31 40
42,020 29,375 26,520 14,280 19,188 16,456 14,850 13,330 16,400
583,696 390,625 270,400 260,100 242,064 234,256 202,500 184,900 168,100
兺x ⫽ 4685
兺y ⫽ 358
兺xy ⫽ 192,419
兺x2 ⫽ 2,536,641
m⫽ ⫽
y
x2
x
60 50
Stories
13.
40 30 20 x 400
600
800
Height (in feet)
n兺xy ⫺ 共兺x兲共兺y兲 9共192,419兲 ⫺ 共4685兲共358兲 ⫽ n兺x2 ⫺ 共兺x兲2 9共2,536,641兲 ⫺ 共4685兲2 54,541 ⬇ 0.06194 ⬇ 0.062 880,544
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CHAPTER 9

CORRELATION AND REGRESSION
4685 ⫺ 共0.06194兲冢 ⬇ 7.535 冢358 9 冣 9 冣
b ⫽ y ⫺ mx ⫽
y ⫽ 0.062x ⫹ 7.535 ^
(a) y ⫽ 0.062共500兲 ⫹ 7.535 ⬇ 38.535 ⬇ 39 stories ^
(b) y ⫽ 0.062共650兲 ⫹ 7.535 ⬇ 47.835 ⬇ 48 stories ^
(c) It is not meaningful to predict the value of y for x ⫽ 310 because x ⫽ 310 is outside the range of the original data. (d) y ⫽ 0.062共725兲 ⫹ 7.535 ⬇ 52.485 ⬇ 52 stories ^
x
y
xy
x2
3 4 4 5 6 2 3
1100 1300 1500 2100 2600 460 1200
3,300 5,200 6,000 10,500 15,600 920 3,600
9 16 16 25 36 4 9
兺x ⫽ 27
兺y ⫽ 10,260
兺xy ⫽ 45,120
兺x 2 ⫽ 115
m⫽
Vocabulary size
y
14.
2800 2400 2000 1600 1200 800 400 x 1 2 3 4 5 6
Age (in years)
n兺xy ⫺ 共兺x兲共兺y兲 7共45,120兲 ⫺ 共27兲共10,260兲 38,820 ⫽ ⫽ ⫽ 510.789 n兺x2 ⫺ 共兺x兲2 7共115兲 ⫺ 共27兲2 76
b ⫽ y ⫺ mx ⫽
27 ⫺ 共510.789兲冢 冣 ⫽ ⫺504.474 冢10,260 7 冣 7
y ⫽ 510.789x ⫺ 504.474 ^
(a) y ⫽ 510.789共2兲 ⫺ 504.474 ⬇ 517
(b) y ⫽ 510.789共3兲 ⫺ 504.474 ⬇ 1028
^
^
(c) y ⫽ 510.789共6兲 ⫺ 504.474 ⬇ 2560 ^
(d) It is not meaningful to predict the value of y for x ⫽ 12 because x ⫽ 12 is outside the range of the original data. x
y
xy
x2
0 1 2 4 4 5 5 5 6 6 7 7 8
40 41 51 48 64 69 73 75 68 93 84 90 95
0 41 102 192 256 345 365 375 408 558 588 630 760
0 1 4 16 16 25 25 25 36 36 49 49 64
兺x ⫽ 60
兺y ⫽ 891
兺xy ⫽ 4620
兺x2 ⫽ 346
m⫽
y 100
Test score
15.
80 60 40 20 x 2
4
6
8
Hours studying
n兺xy ⫺ 共兺x兲共兺y兲 13共4620兲 ⫺ 共60兲共891兲 6600 ⫽ ⫽ ⬇ 7.350 n兺x2 ⫺ 共兺x兲2 13共346兲 ⫺ 共60兲2 898
b ⫽ y ⫺ mx ⫽
891 ⫺ 共7.350兲冢 ⬇ 34.617 冢60 13 冣 13 冣
y ⫽ 7.350x ⫹ 34.617
^
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CORRELATION AND REGRESSION
(a) y ⫽ 7.350共3兲 ⫹ 34.617 ⬇ 56.7 ^
(b) y ⫽ 7.350共6.5兲 ⫹ 34.617 ⬇ 82.4 ^
(c) It is not meaningful to predict the value of y for x ⫽ 13 because x ⫽ 13 is outside the range of the original data. (d) y ⫽ 7.350共4.5兲 ⫹ 34.617 ⬇ 67.7 ^
16.
y
x
y
xy
x2
0 1 2 3 3 5 5 5 6 7 7 10
96 85 82 74 95 68 76 84 58 65 75 50
0 85 164 222 285 340 380 420 348 455 525 500
0 1 4 9 9 25 25 25 36 49 49 100
兺x ⫽ 54
兺y ⫽ 908
兺xy ⫽ 3724
兺x 2 ⫽ 332
Test score
100 80 60 40 20 x 2
4
6
8
10
Hours online
n兺xy ⫺ 共兺x兲共兺y兲 12共3724兲 ⫺ 共54兲共908兲 ⫺4344 ⫽ ⫽ ⬇ ⫺4.067 n兺x2 ⫺ 共兺x兲2 12共332兲 ⫺ 共54兲2 1068
m⫽
b ⫽ y ⫺ mx ⫽
54 ⫺ 共⫺4.067兲冢 冣 ⬇ 93.970 冢908 12 冣 12
y ⫽ ⫺4.067x ⫹ 93.970 ^
(a) y ⫽ ⫺4.067共4兲 ⫹ 93.970 ⬇ 77.7
(b) y ⫽ ⫺4.067共8兲 ⫹ 93.970 ⬇ 61.4
^
^
(c) y ⫽ ⫺4.067共9兲 ⫹ 93.970 ⬇ 57.4 ^
(d) It is not meaningful to predict the values of y for x ⫽ 15 because x ⫽ 15 is outside the range of the original data. x
y
xy
x2
150 170 120 120 90 180 170 140 90 110
420 470 350 360 270 550 530 460 380 330
63,000 79,900 42,000 43,200 24,300 99,000 90,100 64,400 34,200 36,300
22,500 28,900 14,400 14,400 8,100 32,400 28,900 19,600 8,100 12,100
兺x ⫽ 1340
兺y ⫽ 4120
兺xy ⫽ 576,400
兺x2 ⫽ 189,400
m⫽
Sodium (in milligrams)
y
17.
550 500 450 400 350 300 250 x 80
120
160
Calories
n兺xy ⫺ 共兺x兲共兺y兲 10共576,400兲 ⫺ 共1340兲共4120兲 243,200 ⫽ ⫽ ⬇ 2.471545 ⬇ 2.472 n兺x2 ⫺ 共兺x兲2 10共189,400兲 ⫺ 共1340兲2 98,400
b ⫽ y ⫺ mx ⫽
1340 ⫺ 2.471545冢 ⫽ 80.813 冢4120 冣 10 10 冣
y ⫽ 2.472x ⫹ 80.813 ^
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CHAPTER 9
CORRELATION AND REGRESSION
(a) y ⫽ 2.472共170兲 ⫹ 80.813 ⫽ 501.053 milligrams ^
(b) y ⫽ 2.472共100兲 ⫹ 80.813 ⫽ 328.213 milligrams ^
(c) y ⫽ 2.472共140兲 ⫹ 80.813 ⫽ 426.893 milligrams ^
(d) It is not meaningful to predict the value of y for x ⫽ 210 because x ⫽ 210 is outside the range of the original data. x
y
x2
xy
70 72 75 76 71 73 85 78 77 80 82
8.3 10.5 11.0 11.4 9.2 10.9 14.9 14.0 16.3 18.0 15.8
581 756 825 866.4 653.2 795.7 1266.5 1092 1255.1 1440 1295.6
4900 5184 5625 5776 5041 5329 7225 6084 5929 6400 6724
兺x ⫽ 839
兺y ⫽ 140.3
兺xy ⫽ 10,826.5
兺x2 ⫽ 64,217
Trunk diameter (in inches)
y
18.
18 16 14 12 10 8 x 70
75
80
85
Height (in feet)
n兺xy ⫺ 共兺x兲共兺y兲 11共10,826.5兲 ⫺ 共839兲共140.3兲 1379.8 ⫽ ⫽ ⫽ 0.55953 ⬇ 0.560 n兺x2 ⫺ 共兺x兲2 11共64,217兲 ⫺ 共839兲2 2466
m⫽
b ⫽ y ⫺ mx ⫽
839 ⫺ 共0.55953兲冢 ⫽ ⫺29.922 冢140.3 11 冣 11 冣
y ⫽ 0.560x ⫺ 29.922 ^
(a) y ⫽ 0.560共74兲 ⫺ 29.922 ⫽ 11.52 inches ^
(b) y ⫽ 0.560共81兲 ⫺ 29.922 ⫽ 15.44 inches ^
(c) It is not meaningful to predict y for x ⫽ 95 because x ⫽ 95 is outside the range of the original data. (d) y ⫽ 0.560共79兲 ⫺ 29.922 ⫽ 14.32 inches ^
x
y
xy
x2
8.5 9.0 9.0 9.5 10.0 10.0 10.5 10.5 11.0 11.0 11.0 12.0 12.0 12.5
66.0 68.5 67.5 70.0 70.0 72.0 71.5 69.5 71.5 72.0 73.0 73.5 74.0 74.0
561.00 616.50 607.50 665.00 700.00 720.00 750.75 729.75 786.50 792.00 803.00 882.00 888.00 925.00
72.25 81.00 81.00 90.25 100.00 100.00 110.25 110.25 121.00 121.00 121.00 144.00 144.00 156.25
兺y ⫽ 993.0
兺xy ⫽ 10,427.0
兺x2 ⫽ 1552.3
兺x ⫽ 146.5
m⫽
Height (in inches)
y
19.
74 72 70 68 66 x 8
9 10 11 12 13
Shoe size
n兺xy ⫺ 共兺x兲共兺y兲 14共10,427.0兲 ⫺ 共146.5兲共993.0兲 503.5 ⫽ ⫽ ⬇ 1.870 n兺x2 ⫺ 共兺x兲2 14共1552.3兲 ⫺ 共146.5兲2 269.95
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265
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CHAPTER 9

b ⫽ y ⫺ mx ⫽
CORRELATION AND REGRESSION
146.5 ⫺ 共1.870兲冢 ⬇ 51.360 冢993.0 14 冣 14 冣
y ⫽ 1.870x ⫹ 51.360 ^
(a) y ⫽ 1.870共11.5兲 ⫹ 51.360 ⬇ 72.865 inches ^
(b) y ⫽ 1.870共8.0兲 ⫹ 51.360 ⬇ 66.32 inches ^
(c) It is not meaningful to predict the value of y for x ⫽ 15.5 because x ⫽ 15.5 is outside the range of the original data. (d) y ⫽ 1.870共10.0兲 ⫹ 51.360 ⬇ 70.06 inches ^
20.
x
y
y
x2
xy
14.9 14.5 13.9 14.1 13.9 13.7 14.3 13.9 14.0 14.1
1.49 2.90 xxx 9.87 xxx 12.33 xxx 8.34 7.00 xxx
0.01 0.04 xxx 0.49 xxx 0.81 xxx 0.36 0.25 xxx
兺x ⫽ 4.5
兺y ⫽ 141.3
兺xy ⫽ 62.92
兺x 2 ⫽ 2.61
Hours slept
0.1 0.2 0.4 0.7 0.6 0.9 2.0 0.6 0.5 0.1
15.0 14.8 14.6 14.4 14.2 14.0 13.8 13.6 x 0.2 0.4 0.6 0.8 1.0
Age (in years)
n兺xy ⫺ 共兺x兲共兺y兲 10共62.92兲 ⫺ 共4.5兲共141.3兲 ⫺6.65 ⫽ ⫽ ⬇ ⫺1.137 n兺x2 ⫺ 共兺x兲2 10共2.61兲 ⫺ 共4.5兲2 5.85
m⫽
b ⫽ y ⫺ mx ⫽
4.5 ⫺ 共⫺1.137兲冢 冣 ⬇ 14.642 冢141.3 10 冣 10
y ⫽ ⫺1.137x ⫹ 14.642 ^
(a) y ⫽ ⫺1.137共0.3兲 ⫹ 14.642 ⬇ 14.30 hours ^
(b) It is not meaningful to predict the value of y for x ⫽ 3.9 because x ⫽ 3.9 is outside the range of the original data. (c) y ⫽ ⫺1.137共0.6兲 ⫹ 14.642 ⬇ 13.96 hours ^
(d) y ⫽ ⫺1.137共0.8兲 ⫹ 14.642 ⬇ 13.73 hours ^
y
xy
x2
5720 4050 6130 5000 5010 4270 5500 5550
2.19 1.36 2.58 1.74 1.78 1.69 1.80 1.87
12,527 5508 15,815 8700 8917.8 7216.3 9900 10,379
32,718,400 16,402,500 37,576,900 25,000,000 25,100,100 18,232,900 30,250,000 30,802,500
兺x ⫽ 41,230
兺y ⫽ 15.01
兺xy ⫽ 78,962.8
兺x2 ⫽ 216,083,300
m⫽
3.0 2.5 2.0 1.5 1.0 x 4000 4500 5000 5500 6000 6500
x
Variability in braking distance (in feet)
y
21.
Weight (in pounds)
n兺xy ⫺ 共兺x兲共兺y兲 8共78,962.8兲 ⫺ 共41,230兲共15.01兲 12,833.7 ⫽ ⫽ ⫽ 0.000447 n兺x2 ⫺ 共兺x兲2 8共216,083,300兲 ⫺ 共41,230兲2 28,753,500
b ⫽ y ⫺ mx ⫽
41,230 ⫺ 共0.000447兲冢 ⫽ ⫺0.425 冢15.01 8 冣 8 冣
y ⫽ 0.000447x ⫺ 0.425 ^
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CHAPTER 9

CORRELATION AND REGRESSION
(a) y ⫽ 0.000447共4500兲 ⫺ 0.425 ⫽ 1.587 feet ^
(b) y ⫽ 0.000447共6000兲 ⫺ 0.425 ⫽ 2.257 feet ^
(c) It is not meaningful to predict the value of y for x ⫽ 7500 because x ⫽ 7500 is outside the range of the original data. (d) y ⫽ 0.000447共5750兲 ⫺ 0.425 ⫽ 2.145 feet ^
y
y
x2
xy
5720 4050 6130 5000 5010 4270 5500 5550
3.78 2.43 4.63 2.88 3.25 2.76 3.42 3.51
21,622 9841.5 28,381.9 14,400 16,282.5 11,785.2 18,810 19,480.5
32,718,400 16,402,500 37,576,900 25,000,000 25,100,100 18,232,900 30,250,000 30,802,500
兺x ⫽ 41,230
兺y ⫽ 26.66
兺xy ⫽ 140,603.2
兺x 2 ⫽ 216,083,300
m⫽
5.0 4.5 4.0 3.5 3.0 2.5 2.0 x 4000 4500 5000 5500 6000 6500
x
Variability in braking distance (in feet)
22.
Weight (in pounds)
n兺xy ⫺ 共兺x兲共兺y兲 8共140,603.2兲 ⫺ 共41,230兲共26.66兲 25,632.2 ⫽ ⫽ ⬇ 0.000891502 n兺x2 ⫺ 共兺x兲2 8共216,083,300兲 ⫺ 共41,230兲2 28,753,500
b ⫽ y ⫺ mx ⫽
41,230 ⫺ 共⫺0.000891502兲冢 ⫽ ⫺1.26208 冢26.66 8 冣 8 冣
y ⫽ 0.000892x ⫺ 1.262 ^
(a) y ⫽ 0.000892共4800兲 ⫺ 1.262 ⫽ 3.020 feet
(b) y ⫽ 0.000892共5850兲 ⫺ 1.262 ⫽ 3.956 feet
^
^
(c) y ⫽ 0.000892共4075兲 ⫺ 1.262 ⫽ 2.373 feet ^
(d) It is not meaningful to predict the value of y for x ⫽ 3000 because x ⫽ 3000 is outside the range of the original data. 23. Substitute a value x into the equation of a regression line and solve for y. 24. Prediction values are meaningful only for xvalues in (or close to) the range of the data. 25. Strong positive linear correlation; As the ages of the engineers increase, the salaries of the engineers tend to increase. 26. m ⫽ 0.510
y ⫽ mx ⫹ b ⫽ 0.510x ⫹ 45.642 ^
Annual salary (in thousands)
b ⫽ 45.642
y 85 80 75 70 65 60 55 50 x 20 30 40 50 60 70
Age (in years)
27. It is not meaningful to predict y for x ⫽ 74 because x ⫽ 74 is outside the range of the original data. 28. r ⫽ 0.917 cv ⫽ 0.684 ⇒ Reject H0: ⫽ 0 29. In general, as age increases; salary increases until age 61 when salary decreases. (Answers will vary.) © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
267
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CHAPTER 9

CORRELATION AND REGRESSION
30. (a) y ⫽ 1.724x ⫹ 79.733
(b) y ⫽ 0.453x ⫺ 26.448
^
^
y
y
150
Row 1
Row 2
200
100 50
70 60 50 40 30 20 10 x
x 50
10 20 30 40 50 60 70
100 150 200
Row 2
Row 1
(c) The slope of the line keeps the same sign, but the values of m and b change. y
31. (a)
y
(b)
100
Row 1
Row 2
90 80 70 60
7 6 5 4 3 2 1
x
x
1 2 3 4 5 6 7
60
70
Row 1
80
90 100
Row 2
y ⫽ ⫺4.297x ⫹ 94.200
y ⫽ ⫺0.1413x ⫹ 14.763
^
^
(c) The slope of the line keeps the same sign, but the values of m and b change. 32. (a) m ⫽ 1.711
y
(b) 30
b ⫽ 3.912
25
y ⫽ 1.711x ⫹ 3.912
20
^
15 10 5 x 2
(c)
4
x
y
^ y ⴝ 1.711x ⴙ 3.912
yⴚ^ y
8 4 15 7 6 3 12 10 5
18 11 29 18 14 8 25 20 12
17.600 10.756 29.577 15.889 14.178 9.045 24.444 21.022 12.467
0.400 0.244 ⫺0.577 2.111 ⫺0.178 ⫺1.045 0.556 ⫺1.022 ⫺0.467
6
8 10 12 14 16
Residual 3 2 1 x 2 4
8 10 12
16
−1 −2
(d) The residual plot shows no pattern in the residuals because the residuals fluctuate about 0. This suggests that the regression line is a good representation of the data. 33. (a) m ⫽ 0.139 b ⫽ 21.024 y ⫽ 0.139x ⫹ 21.024 ^
y
(b) 32 30 28 26 24 22
x 30
40
50
60
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CHAPTER 9
(c)

CORRELATION AND REGRESSION
Residual
x
y
^ y ⴝ 0.139x ⴙ 21.024
yⴚ^ y
38 34 40 46 43 48 60 55 52
24 22 27 32 30 31 27 26 28
26.306 25.750 26.584 27.418 27.001 27.696 29.364 28.669 28.252
⫺2.306 ⫺3.750 0.416 4.582 2.999 3.304 ⫺2.364 ⫺2.669 ⫺0.252
6 4 2 x 30
40
50
60
−2 −4
(d) The regression line may not be a good model for the data because the data do not fluctuate about 0. y
34. (a) 15 12 9 6 3
x 3
6
9
12
15
(b) The point 共14, 3兲 may be an outlier. (c) The point 共14, 3兲 is influential because using all 6 points ⇒ y ⫽ 0.212x ⫹ 6.445. ^
However, excluding the point 共14, 3兲 ⇒ y ⫽ 0.905 ⫹ 3.568. ^
y
35. (a) 35 30 25 20 15 10 5
x 10
20
30
40
50
(b) The point 共44, 8兲 may be an outlier. (c) Excluding the point 共44, 8兲⇒ y ⫽ ⫺0.711x ⫹ 35.263. The point 共44, 8兲 is not influential because using all 8 points ⇒ y ⫽ ⫺0.607x ⫹ 34.160. ^
^
36. The point 共200, 1835.4兲 is an outlier, but is not an influential point because the regression line shows no significant change. 37. m ⫽ 654.536
y ⫽ 654.536x ⫺ 1214.857
^
Number of bacteria
b ⫽ ⫺1214.857
y 5000 4000 3000 2000 1000 x 1 2 3 4 5 6 7
Number of hours
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38.

CORRELATION AND REGRESSION
x
y
log y
1 2 3 4 5 6 7
165 280 468 780 1310 1920 4900
2.218 2.447 2.670 2.892 3.117 3.283 3.690
log y 4 3 2 1 x 1
2
3
4
5
6
7
log y ⫽ 0.233x ⫹ 1.969 39. log y ⫽ 0.233x ⫹ 1.969 ⇒ y ⫽ 10 0.233x ⫹1.969 ⇒ y ⫽ 93.111共10兲 0.233x 40. The exponential equation is a better model for the data. The exponential equation’s error is smaller. 41. m ⫽ ⫺78.929
y 700
b ⫽ 576.179
600
y ⫽ ⫺78.929x ⫹ 576.179
500
^
400 300 200 100 x 1 2 3 4 5 6 7 8
42.
x
y
log x
log y
1 2 3 4 5 6 7 8
695 410 256 110 80 75 68 74
0 0.301 0.477 0.602 0.699 0.778 0.845 0.903
2.842 2.613 2.408 2.041 1.903 1.875 1.833 1.869
log y ⫽ m共log x兲 ⫹ b ⇒ log y ⫽ ⫺1.251 log x ⫹ 2.893 A linear model is more appropriate for the transformed data.
log y 3.0 2.8 2.6 2.4 2.2 2.0 1.8 log x 0.2 0.4 0.6 0.8
1
43. log y ⫽ m共log x兲 ⫹ b ⇒ y ⫽ 10m log x ⫹ b ⫽ 10 b 10 m log x ⫽ 102.893 10⫺1.251 log x ⫽ 781.628
⭈ x⫺1.251
44. The power equation is a much better model for the data. The power equation’s error is smaller.
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CHAPTER 9
45. y ⫽ a ⫹ b ln x ⫽ 25.035 ⫹ 19.599 ln x
CORRELATION AND REGRESSION
46. y ⫽ a ⫹ b ln x ⫽ 13.667 ⫺ 0.471 ln x ^
16
76
Height (in inches)

8
0
13
1 12
64
Shoe size
47. The logarithmic equation is a better model for the data. The logarithmic equation’s error is smaller. 48. The logarithmic equation is a better model for the data. The logarithmic equation’s error is smaller.
9.3 MEASURES OF REGRESSION AND PREDICTION INTERVALS
9.3 Try It Yourself Solutions b. r 2 ⫽ 共0.979兲2 ⫽ 0.958
1a. r ⫽ 0.979
c. 95.8% of the variation in the times is explained. 4.2% of the variation is unexplained. 2a.
xi
yi
^ yi
共 yi ⴚ ^ yi 兲2
15 20 20 30 40 45 50 60
26 32 38 56 54 78 80 88
28.386 35.411 35.411 49.461 63.511 70.536 77.561 91.611
5.693 11.635 6.703 42.759 90.459 55.711 5.949 13.039 兺 ⫽ 231.948
b. n ⫽ 8 ⬇ 6.218 冪兺共 ny ⫺⫺ 2y 兲 ⫽ 冪231.948 6 ^
c. se ⫽
i
i
2
d. The standard error of estimate of the weekly sales for a specific radio ad time is about $621.80. 3a. n ⫽ 8, d.f. ⫽ 6, tc ⫽ 2.447, se ⬇ 10.290 b. y ⫽ 50.729x ⫹ 104.061 ⫽ 50.729共2.5兲 ⫹ 104.061 ⬇ 230.884 ^
冪1 ⫹ n1 ⫹ n共兺xn共x兲 ⫺⫺ x共兲兺x兲
c. E ⫽ tc se
2
2
2
共2.5 ⫺ 1.975兲 冪1 ⫹ 18 ⫹ 8共832.44 兲 ⫺ 共15.8兲
⫽ 共2.447兲共10.290兲
2
2
⫽ 共2.447兲共10.290兲冪1.34818 ⬇ 29.236 d. y ± E → 共201.648, 260.120兲 ^
e. You can be 95% confident that the company sales will be between $201,648 and $260,120 when advertising expenditures are $2500. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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9.3 EXERCISE SOLUTIONS 1. Total variation ⫽ 兺共 yi ⫺ y 兲2 ; the sum of the squares of the differences between the yvalues of each ordered pair and the mean of the yvalues of the ordered pairs. 2. Explained variation ⫽ 兺共 yi ⫺ y兲2; the sum of the squares of the differences between the predicted yvalues and the mean of the yvalues of the ordered pairs. ^
3. Unexplained variation ⫽ 兺共 yi ⫺ yi 兲2 ; the sum of the squares of the differences between the observed yvalues and the predicted yvalues. ^
兺共 yi ⫺ y兲2 兺共 yi ⫺ y兲2 ^
4. Coefficient of determination: r 2 ⫽
r 2 is the ratio of the explained variation to the total variation and is the percent of variation of y that is explained by the relationship between x and y. 5. r 2 ⫽ 共0.350兲2 ⬇ 0.123 12.3% of the variation is explained. 87.7% of the variation is unexplained. 6. r 2 ⫽ 共⫺0.275兲2 ⬇ 0.076 7.6% of the variation is explained. 92.4% of the variation is unexplained. 7. r ⫽ 共⫺0.891兲2 ⬇ 0.794 79.4% of the variation is explained. 20.6% of the variation is unexplained. 8. r 2 ⫽ 共0.964兲2 ⬇ 0.929 92.9% of the variation is explained. 7.1% of the variation is unexplained. 兺共 yi ⫺ y兲2 ⬇ 0.233 兺共 yi ⫺ y 兲2 ^
9. (a) r 2 ⫽
23.3% of the variation in proceeds can be explained by the variation in the number of issues and 76.7% of the variation is unexplained. (b) se ⫽
冪
兺共 yi ⫺ yi 兲2 ⫽ n⫺2 ^
⬇ 15,349.725 冪2,356,140,670 10
The standard error of estimate of the proceeds for a specific number of issues is about $15,349,725,000. 兺共 yi ⫺ y兲2 ⬇ 0.787 兺共 yi ⫺ y兲2 ^
10. (a) r 2 ⫽
78.7% of the variation in cigarettes exported from the United States can be explained by the variation in the number of cigarettes consumed in the United States and 21.3% of the variation is unexplained. (b) se ⫽
冪
兺共 yi ⫺ yi 兲2 ⫽ n⫺2 ^
⬇ 8.574 冪367.604 5
The standard error of estimate of the number of cigarettes exported for a specific number of cigarettes consumed is about 8,574,000,000. 兺共 yi ⫺ y兲2 ⬇ 0.981 兺共 yi ⫺ y兲2 ^
11. (a) r 2 ⫽
98.1% of the variation in sales can be explained by the variation in the total square footage and 1.9% of the variation is unexplained.
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CHAPTER 9

CORRELATION AND REGRESSION
⬇ 30.576 冪兺共 ny ⫺⫺ 2y 兲 ⫽ 冪8413.958 9 ^
(b) se ⫽
i
i
2
The standard error of estimate of the sales for a specific total square footage is about 30,576,000,000. 兺共 yi ⫺ y兲2 ⬇ 0.578 兺共 yi ⫺ y兲2 ^
12. (a) r 2 ⫽
57.8% of the variation in the median number of leisure hours per week can be explained by the variation in the median number of work hours per week and 42.2% of the variation is unexplained. ⬇ 2.013 冪兺共 ny ⫺⫺ 2y 兲 ⫽ 冪33.218 8 ^
(b) se ⫽
i
i
2
The standard error of estimate of the median number of leisure hours/week for a specific median number of hours/week is about 2.013 hours. 兺共 yi ⫺ y兲2 ⬇ 0.994 兺共 yi ⫺ y兲2 ^
13. (a) r 2 ⫽
99.4% of the variation in the median weekly earnings of female workers can be explained by the variation in the median weekly earnings of male workers and 0.6% of the variation is unexplained. ⬇ 2.719 冪兺共 ny ⫺⫺ 2y 兲 ⫽ 冪22.178 3 ^
(b) se ⫽
i
i
2
The standard error of estimate of the median weekly earnings of female workers for a specific median weekly earnings of male workers is about $2.72. 兺共 yi ⫺ y兲2 ⬇ 0.931 兺共 yi ⫺ y兲2 ^
14. (a) r 2 ⫽
93.1% of the variation in the turnout for federal elections can be explained by the variation in the voting age population and 6.9% of the variation is unexplained. ⬇ 2.296 冪兺共 ny ⫺⫺ 2y 兲 ⫽ 冪31.619 6 ^
(b) se ⫽
i
i
2
The standard error of estimate of the turnout in a federal election for a specified voting age population is about 2,296,000. 兺共 yi ⫺ y兲2 ⬇ 0.994 兺共 yi ⫺ y兲2 ^
15. (a) r 2 ⫽
99.4% of the variation in the money spent can be explained by the variation in the money raised and 0.6% of the variation is unexplained. (b) se ⫽
冪
兺共 yi ⫺ yi 兲2 ⫽ n⫺2 ^
⬇ 18.869 冪2136.181 6
The standard error of estimate of the money spent for a specified amount of money raised is about $18,869,000. 兺共 yi ⫺ y兲2 ⬇ 0.905 兺共 yi ⫺ y兲2 ^
16. (a) r 2 ⫽
90.5% of the variation in federal pension plans can be explained by the variation in IRA’s and 9.5% of the variation is unexplained. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
273
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CHAPTER 9

CORRELATION AND REGRESSION
⬇ 44.987 冪兺共 ny ⫺⫺ 2y 兲 ⫽ 冪14,166.620 7 ^
(b) se ⫽
i
i
2
The standard error of estimate of assets in federal pension plans for a specified total of IRA assets is about $44,987,000,000. 17. n ⫽ 12, d.f. ⫽ 10, tc ⫽ 2.228, se ⫽ 15,349.725 y ⫽ 40.049x ⫹ 21,843.09 ⫽ 40.049共712兲 ⫹ 21,843.09 ⬇ 50,357.978 ^
冪1 ⫹ n1 ⫹ n共兺xn共x兲 ⫺⫺ x共兲兺x兲 2
E ⫽ tc se
2
2
冪
⫽ 共2.228兲共15,349.725兲
2
1⫹
1 12共712 ⫺ 3806兾12兲 ⫹ 12 12共1,652,648兲 ⫺ 共3806兲2
⫽ 共2.228兲共15,349.725兲冪1.43325 ⬇ 40,942.727 y ± E → 共9415.251, 91,300.705兲 → 共$9,415,251,000, $91,300,705,000兲 ^
You can be 95% confident that the proceeds will be between $9,415,251,000 and $91,300,705,000 when the number of initial offerings is 712. 18. n ⫽ 7, d.f. ⫽ 5, tc ⫽ 2.571, se ⫽ 8.574 y ⫽ 0.671x ⫹ 136.771 ⫽ 0.671共450兲 ⫺ 136.771 ⬇ 165.179 ^
冪1 ⫹ n1 ⫹ n共兺xn共x兲 ⫺⫺ x共兲兺x兲 2
E ⫽ tc se
2
2
关450 ⫺ 共2869兾7兲兴 冪1 ⫹ 17 ⫹ 7共71,178,895 兲 ⫺ 共2869兲
⫽ 共2.571兲共8.574兲
2
2
⫽ 共2.571兲共8.574兲 冪1.67736 ⬇ 28.550 y ± E → 共136.629, 193.729兲 → 共136,629,000,000, 193,729,000,000兲 ^
You can be 95% confident that the number of cigarettes exported will be between 136,629,000,000 and 193,729,000,000 when the number of cigarettes consumed is 450 billion. 19. n ⫽ 11, d.f. ⫽ 9, tc ⫽ 1.833, se ⬇ 30.576 y ⫽ 548.448x ⫺ 1881.694 ⫽ 548.448共4.5兲 ⫺ 1881.694 ⬇ 590.822 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x兲2 ⫹ ⫽ 共1.833兲共30.576兲 n n共兺x2兲 ⫺ 共兺x兲2
冪
1⫹
2
1 11关4.5 ⫺ 共61.2兾11兲兴 ⫹ 11 11共341.9兲 ⫺ 共61.2兲2
⫽ 共1.833兲共30.576兲冪1.89586 ⬇ 77.170 y ± E → 共513.652, 677.992兲 → 共$513,652,000,000, $667,992,000,000兲 ^
You can be 90% confident that the sales will be between $513,652,000,000 and $667,992,000,000 when the total square footage is 4.5 billion.
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CHAPTER 9

CORRELATION AND REGRESSION
20. n ⫽ 10, d.f. ⫽ 8, tc ⫽ 1.860, se ⬇ 2.013 y ⫽ ⫺0.646x ⫹ 50.734 ⫽ ⫺0.646共45.1兲 ⫹ 50.734 ⬇ 21.599 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x兲2 ⫹ n n共兺x2兲 ⫺ 共兺x兲2
关45.1 ⫺ 共475.5兾10兲兴 冪1 ⫹ 101 ⫹ 1010共22,716.29 兲 ⫺ 共475.5兲 2
⫽ 共1.860兲共2.013兲
2
⫽ 共1.860兲共2.013兲 冪1.15649 ⬇ 4.026 y ± E → 共17.573, 25.625兲 ^
You can be 90% confident that the median number of leisure hours/week will be between 17.573 and 25.625 when the median number of work hours/week is 45.1. 21. n ⫽ 5, d.f. ⫽ 3, tc ⫽ 5.841, se ⬇ 2.719 y ⫽ 1.369x ⫺ 402.687 ⫽ 1.369共650兲 ⫺ 402.687 ⬇ 487.163 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x兲2 ⫹ ⫽ 共5.841兲共2.719兲 n n共兺x2兲 ⫺ 共兺x兲2
冪
1⫹
2
1 5关650 ⫺ 共3479兾5兲兴 ⫹ 5 5共2,422,619兲 ⫺ 共3479兲2
⫽ 共5.841兲共2.719兲冪2.28641 ⬇ 24.014 y ± E → 共463.149, 511.177兲 ^
You can be 99% confident that the median earnings of female workers will be between $463.15 and $511.18 when the median weekly earnings of male workers is $650. 22. n ⫽ 8, d.f. ⫽ 6, tc ⫽ 3.707, se ⬇ 2.296 y ⫽ 0.342x ⫹ 5.990 ⫽ 0.342共210兲 ⫹ 5.990 ⬇ 77.81 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x兲2 ⫹ n n共兺x2兲 ⫺ 共兺x兲2
冪
⫽ 共3.707兲共2.296兲
1⫹
1 8关210 ⫺ 共1449.1兾8兲兴 2 ⫹ 8 8共266,100.61兲 ⫺ 共1449.1兲2
⫽ 共3.707兲共2.296兲 冪1.35549 ⬇ 9.909 y ± E → 共67.901, 87.719兲 ^
You can be 99% confident that the voter turnout in federal elections will be between 67.901 million and 87.719 million when the voting age population is 210 million. 23. n ⫽ 8, d.f. ⫽ 6, tc ⫽ 2.447, se ⬇ 18.869 y ⫽ 0.943x ⫺ 21.541 ⫽ 0.943共775.8兲 ⫹ 21.541 ⬇ 753.120 ^
冪1 ⫹ n1 ⫹ n共兺xn共x兲 ⫺⫺ x共兲兺x兲
E ⫽ tc se
2
2
2
8关775.8 ⫺ 共6666.2兾8兲兴 冪1 ⫹ 18 ⫹ 8共5,932,282.32 兲 ⫺ 共6666.2兲
⫽ 共2.447兲共18.869兲
2
2
⫽ 共2.447兲共18.869兲冪1.13375 ⬇ 49.163 y ± E → ($703.957 million, $802.283 million) ^
You can be 95% confident that the money spent in congressional campaigns will be between $703.957 million and $802.283 million when the money raised is $775.8 million.
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275
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CORRELATION AND REGRESSION
24. n ⫽ 9, d.f. ⫽ 7, tc ⫽ 1.895, se ⬇ 44.987 y ⫽ 0.21x ⫹ 288.18 ⫽ 0.21共2500兲 ⫹ 288.18 ⬇ 813.18 ^
1 n共x ⫺ x兲2 ⫹ n n共兺x2兲 ⫺ 共兺x兲2
冪
E ⫽ tc se
1⫹
关2500 ⫺ 共22,332兾9兲兴 冪1 ⫹ 91 ⫹ 9共958,537,290 兲 ⫺ 共22,332兲 2
⫽ 共1.895兲共44.987兲
2
⫽ 共1.895兲共44.987兲 冪1.11122 ⬇ 89.866 y ± E → 共723.314, 903.046兲 ^
You can be 90% confident that the total assets in federal pension plans will be between $723.314 billion and $903.046 billion when the total assets in IRA’s is $2500 billion. 26. y ⫽ 0.217x ⫹ 4.966
y
25.
^
y = 6.586
7.0
y
y = 6.586
7.0
6.0
x = 7.471
Trucks
Trucks
6.5
5.5
6.5 6.0
x = 7.471
x 4
5
6
7
8
5.5
9
Cars
x 4
5
6
7
8
9
Cars
27.
xi
yi
^ yi
^ yi ⴚ y
yi ⴚ ^ yi
yi ⴚ y
9.2 9.0 8.4 8.3 6.5 6.0 4.9
6.9 6.8 6.8 6.9 6.5 6.3 5.9
6.962 6.919 6.789 6.767 6.377 6.268 6.029
0.376 0.333 0.203 0.181 ⫺ 0.210 ⫺ 0.318 ⫺ 0.557
⫺ 0.062 ⫺0.119 0.011 0.133 0.124 0.032 ⫺0.129
0.314 0.214 0.214 0.314 ⫺0.086 ⫺0.286 ⫺0.686
28. (a) Explained variation ⫽ 兺共 yi ⫺ y兲2 ⬇ 0.781 ^
(b) Unexplained variation ⫽ 兺共 yi ⫺ yi 兲2 ⬇ 0.069 ^
(c) Total variation ⫽ 兺共 yi ⫺ y兲2 ⬇ 0.849 29. r 2 ⬇ 0.919; About 91.9% of the variation in the median age of trucks can be explained by the variation in the median age of cars, and 8.1% of the variation is unexplained. 30. se ⬇ 0.117; The standard error of estimate of the median age of trucks for a specific median age of cars is about 0.117 year. 31. y ⫽ 0.217x ⫹ 4.966 ⫽ 0.217共8.6兲 ⫹ 4.966 ⬇ 6.832 ^
冪1 ⫹ n1 ⫹ n共兺xn共x兲 ⫺⫺ x共兲兺x兲
E ⫽ tc se
2
2
2
8.6 ⫺ 共52.3兾7兲兴 冪1 ⫹ 17 ⫹ 77共关407.35 兲 ⫺ 共52.3兲
⫽ 共2.571兲共0.117兲
2 2
⫽ 共2.571兲共0.117兲冪1.21961 ⬇ 0.332 y ± E → 共6.500, 7.164兲 ^
32. The slope and correlation coefficient will have the same sign because the denominators of both formulas are always positive while the numerators are always equal.
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CHAPTER 9

CORRELATION AND REGRESSION
33. cv ⫽ ± 3.707 m ⫽ ⫺0.205 se ⫽ 0.554 t⫽
m se
冪兺x
2
⫺
(兺x兲2 ⫺0.205 ⫽ n 0.554
冪838.55 ⫺ 共79.58 兲
2
⫽ ⫺2.578
Fail to reject H0: M ⫽ 0 34. cv ⫽ ± 2.306 m ⫽ 1.328 se ⫽ 5.086 t⫽
m se
冪兺x
2
⫺
(兺x兲2 1.328 ⫽ n 5.086
冪14,660 ⫺ 共37410 兲
2
⫽ 6.771
Reject H0: M ⫽ 0
冪
35. E ⫽ tc se
冪
1 x2 ⫹ 2 ⫽ 2.447共10.290兲 n 兺x ⫺ [共兺x兲2兾n兴
1 共15.8 8 兲 ⫹ ⫽ 45.626 8 共32.44兲 ⫺ 共15.8兲2 2
8
b ± E ⇒ 104.061 ± 45.626 ⫽ 共58.435, 149.687兲 E⫽
tc se
冪
兺x2
共兺x兲 ⫺ n
2
⫽
2.447共10.290兲
冪
共15.8兲2 32.44 ⫺ 8
⫽ 22.658
m ± E ⇒ 50.729 ± 22.658 ⇒ 共28.071, 73.387兲
冪
36. E ⫽ tc se
冪
1 x2 ⫹ 2 ⫽ 共3.707兲共10.290兲 n 兺x ⫺ 共兺x兲2
1 共15.8 8 兲 ⫹ ⫽ 69.119 8 共32.44兲 ⫺ 共15.8兲2 2
8
b ± E ⇒ 104.061 ± 69.119 ⇒ 共34.942, 173.18兲 E⫽
tcse
冪
兺x2
共兺x兲2 ⫺ n
⫽
共3.707兲共10.290兲
冪
共15.8兲2 32.44 ⫺ 8
⫽ 34.325
m ± E ⇒ 50.729 ± 34.325 ⇒ 共16.404, 85.054兲
9.4 MULTIPLE REGRESSION
9.4 Try It Yourself Solutions 1a. Enter the data.
b. y ⫽ 46.385 ⫹ 0.540x1 ⫺ 4.897x 2 ^
2ab. (1) y ⫽ 46.385 ⫹ 0.540共89兲 ⫺ 4.897共1兲 ^
(2) y ⫽ 46.385 ⫹ 0.540共78兲 ⫺ 4.897共3兲 ^
(3) y ⫽ 46.385 ⫹ 0.540共83兲 ⫺ 4.897共2兲 ^
c. (1) y ⫽ 89.548 ^
d. (1) 90
(2) 74
(2) y ⫽ 73.814 ^
(3) y ⫽ 81.411 ^
(3) 81
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9.4 EXERCISE SOLUTIONS 1. y ⫽ 640 ⫺ 0.105x1 ⫹ 0.124x2 ^
(a) y ⫽ 640 ⫺ 0.105共13,500兲 ⫹ 0.124共12,000兲 ⫽ 710.5 pounds ^
(b) y ⫽ 640 ⫺ 0.105共15,000兲 ⫹ 0.124共13,500兲 ⫽ 739 pounds ^
(c) y ⫽ 640 ⫺ 0.105共14,000兲 ⫹ 0.124共13,500兲 ⫽ 844 pounds ^
(d) y ⫽ 640 ⫺ 0.105共14,500兲 ⫹ 0.124共13,000兲 ⫽ 729.5 pounds ^
2. y ⫽ 17 ⫹ ⫺0.000642x 1 ⫹ 0.00125x 2 ^
(a) y ⫽ 17 ⫹ ⫺0.000642共60,000兲 ⫹ 0.00125共45,000兲 ⫽ 34.73 bushels per acre ^
(b) y ⫽ 17 ⫹ ⫺0.000642共62,500兲 ⫹ 0.00125共53,000兲 ⫽ 43.125 bushels per acre ^
(c) y ⫽ 17 ⫹ ⫺0.000642共57,500兲 ⫹ 0.00125共50,000兲 ⫽ 42.585 bushels per acre ^
(d) y ⫽ 17 ⫹ ⫺0.000642共59,000兲 ⫹ 0.00125共48,500兲 ⫽ 39.747 bushels per acre ^
3. y ⫽ ⫺52.2 ⫹ 0.3x1 ⫹ 4.5x2 ^
(a) y ⫽ ⫺52.2 ⫹ 0.3共70兲 ⫹ 4.5共8.6兲 ⫽ 7.5 cubic feet ^
(b) y ⫽ ⫺52.2 ⫹ 0.3共65兲 ⫹ 4.5共11.0兲 ⫽ 16.8 cubic feet ^
(c) y ⫽ ⫺52.2 ⫹ 0.3共83兲 ⫹ 4.5共17.6兲 ⫽ 51.9 cubic feet ^
(d) y ⫽ ⫺52.2 ⫹ 0.3共87兲 ⫹ 4.5共19.6兲 ⫽ 62.1 cubic feet ^
4. y ⫽ ⫺0.415 ⫹ 0.0009x1 ⫹ 0.168x2 ^
(a) y ⫽ ⫺0.415 ⫹ 0.0009共15.2兲 ⫹ 0.168共10.3兲 ⫽ $1.33 ^
(b) y ⫽ ⫺0.415 ⫹ 0.0009共19.8兲 ⫹ 0.168共14.5兲 ⫽ $2.04 ^
(c) y ⫽ ⫺0.415 ⫹ 0.0009共12.1兲 ⫹ 0.168共9.1兲 ⫽ $1.12 ^
(d) y ⫽ ⫺0.415 ⫹ 0.0009共21.5兲 ⫹ 0.168共15.8兲 ⫽ $2.26 ^
5. y ⫽ ⫺2518.4 ⫹ 126.8x1 ⫹ 66.4x2 ^
(a) s ⫽ 28.489 (b) r 2 ⫽ 0.985 (c) The standard error of estimate of the predicted sales given a specific total square footage and number of shopping centers is $28.489 billion. The multiple regression model explains 98.5% of the variation in y. 6. y ⫽ ⫺4.312 ⫹ 0.2366x1 ⫺ 0.1093x2 ^
(a) s ⫽ 0.828
(b) r 2 ⫽ 0.996
(c) The standard error of estimate of the predicted equity given a specific net sales and total assets is $0.828 billion. The multiple regression model explains 99.6% of the variation in y.
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CHAPTER 9

CORRELATION AND REGRESSION
7. n ⫽ 11, k ⫽ 2, r 2 ⫽ 0.985 2 ⫽ 1 ⫺ radj
冤 共1 n⫺⫺r k兲共n⫺⫺1 1兲冥 ⫽ 0.981 2
98.1% of the variation in y can be explained by the relationships between variables. 2 < r2 radj
8. n ⫽ 6, k ⫽ 2, r 2 ⫽ 0.996 2 ⫽ 1 ⫺ radj
冤
共1 ⫺ r 2兲共n ⫺ 1兲 ⫽ 0.993 n⫺k⫺1
冥
99.3% of the variation in y can be explained by the relationships between variables. 2 < r2 radj
CHAPTER 9 REVIEW EXERCISE SOLUTIONS y
y
2.
38
Average monthly temperature (in °F)
Milk production (in gallons)
1. 36 34 32 30
100 80 60 40 20
x
x
4 5 6 7 8 9 10 11
2
Age (in years)
y
8 10
r ⬇ 0.944; strong positive linear correlation; The average monthly temperatures increase with convertibles sold. y
4.
1000
2.5
950
2.0
Cavities
Brain size (pixel count in thousands)
6
Convertibles sold
r ⬇ ⫺0.939; strong negative linear correlation; milk production decreases as age increases
3.
4
900 850 800
1.5 1.0 0.5
750
x
x 60
100
2
140
6
8 10 12
Sugar consumption (in kilograms)
IQ
r ⬇ 0.361; weak positive linear correlation; brain size increases as IQ increases.
4
r ⬇ 0.953; strong positive linear correlation; The number of cavities increases as the annual per capita sugar consumption increases.
5. H0: ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.01, d.f. ⫽ n ⫺ 2 ⫽ 24 t0 ⫽ 2.797 r t⫽ ⫽ 1 ⫺ r2 n⫺2
冪
0.24
冪
1 ⫺ 共0.24兲 26 ⫺ 2
2
⫽
0.24 冪0.03927
⫽ 1.211
Fail to reject H0 . There is not enough evidence to conclude that a significant linear correlation exists. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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CORRELATION AND REGRESSION
6. H0: ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.05, d.f. ⫽ n ⫺ 2 ⫽ 20 t0 ⫽ ± 2.086 t⫽
r
冪
⫽
1⫺ n⫺2
r2
⫺0.55
冪
1 ⫺ 共⫺0.55兲 22 ⫺ 2
2
⫽
⫺0.55 冪0.03488
⬇ ⫺2.945
Reject H0. There is enough evidence to conclude that a significant linear correlation exists. 7. H0: ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.05, d.f. ⫽ n ⫺ 2 ⫽ 6 t0 ⫽ ± 2.447 r t⫽ ⫽ 1 ⫺ r2 n⫺2
冪
⫺0.939
冪
1 ⫺ 共⫺0.939兲 8⫺2
2
⫽
⫺0.939 冪0.01971
⫽ ⫺6.688
Reject H0 . There is enough evidence to conclude that there is a significant linear correlation between the age of a cow and its milk production. 8. H0: ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.05, d.f. ⫽ n ⫺ 2 ⫽ 4 t0 ⫽ ± 2.776 t⫽
r
冪
1 ⫺ r2 n⫺2
⫽
0.944
冪
1 ⫺ 共0.944兲2 4
⫽ 5.722
Reject H0. There is enough evidence to conclude that a significant linear correlation exists between average monthly temperatures and convertible sales. 9. H0: ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.01, d.f. ⫽ n ⫺ 2 ⫽ 7 t0 ⫽ ± 3.499 r t⫽ ⫽ 1 ⫺ r2 n⫺2
冪
0.361
冪
1 ⫺ 共0.361兲2 7
⫽ 1.024
Fail to reject H0. There is not enough evidence to conclude a significant linear correlation exists between brain size and IQ. 10. H0: ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.01, d.f. ⫽ n ⫺ 2 ⫽ 5 t0 ⫽ ± 4.032 t⫽
r
⫽
0.953
兲 冪1n ⫺⫺ r2 冪1 ⫺7共⫺0.953 2 2
2
⫽
0.953 ⬇ 7.034 冪0.01836
Reject H0. There is enough evidence to conclude that there is a linear correlation between sugar consumption and tooth decay.
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CHAPTER 9
11. y ⫽ 0.757x ⫹ 21.525
^
y
y 74
Time (in minutes)
Brother height (in inches)
CORRELATION AND REGRESSION
12. y ⫽ 3.431x ⫹ 193.845
^
72 70 68 66
500 490 480 470 460 450 440 430
x 60
62
64
66
x
68
72 76 80 84 88
Sister height (in inches)
Households (in millions)
r ⬇ 0.688 (Moderate positive linear correlation) 13. y ⫽ ⫺0.086x ⫹ 10.450 ^
r ⬇ 0.970 (Strong positive linear correlation) 14. y ⫽ ⫺0.090x ⫹ 44.675 ^
y
8 7 6 5 4 x
Fuel efficiency (in miles per gallon)
y
9
Hours of sleep

35 30 25 20 15 10 5 x
20 30 40 50 60 70 80
100 150 200 250 300 350
Age (in years)
r ⬇ ⫺0.949 (Strong negative linear correlation)
Displacement (in cubic inches)
r ⬇ ⫺0.984 (Strong negative linear correlation)
15. (a) y ⫽ 0.757共61兲 ⫹ 21.525 ⫽ 67.702 ⬇ 68 inches ^
(b) y ⫽ 0.757共66兲 ⫹ 21.525 ⫽ 71.487 ⬇ 71 inches ^
(c) Not meaningful because x ⫽ 75 inches is outside range of data. (d) Not meaningful because x ⫽ 50 inches is outside range of data. 16. (a) Not meaningful because x ⫽ 60 is outside the range of the original data. (b) y ⫽ 3.431共85兲 ⫹ 193.845 ⫽ 485.48 minutes ^
(c) y ⫽ 3.431共75兲 ⫹ 193.845 ⫽ 451.17 minutes ^
(d) Not meaningful because x ⫽ 97 is outside the range of the original data. 17. (a) 8.902 hours (b) y ⫽ 10.450 ⫽ 0.086x ⫽ 10.450 ⫺ 0.086共25兲 ⫽ 8.3 hours ^
(c) Not meaningful because x ⫽ 85 years is outside range of data. (d) y ⫽ 10.450 ⫺ 0.086x ⫽ 10.450 ⫺ 0.086共50兲 ⫽ 6.15 hours ^
18. (a) Not meaningful because x ⫽ 86 is outside the range of the original data. (b) y ⫽ ⫺0.090共198兲 ⫹ 44.675 ⫽ 26.86 miles per gallon ^
(c) y ⫽ ⫺0.090共289兲 ⫹ 44.675 ⫽ 18.67 miles per gallon ^
(d) Not meaningful because x ⫽ 407 is outside the range of the original data. 19. r 2 ⫽ 共⫺0.553兲2 ⫽ 0.306 30.6% of the variation in y is explained. 69.4% of the variation in y is unexplained. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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20. r 2 ⫽ 共⫺0.962兲2 ⫽ 0.925 92.5% of the variation in y is explained. 7.5% of the variation in y is unexplained. 21. r 2 ⫽ 共0.181兲2 ⫽ 0.033 3.3% of the variation in y is explained. 96.7% of the variation in y is unexplained. 22. r 2 ⫽ 共0.740兲2 ⫽ 0.548 54.8% of the variation in y is explained. 45.2% of the variation in y is unexplained. 23. (a) r 2 ⫽ 0.897 89.7% of the variation in y is explained. 10.3% of the variation in y is unexplained. (b) se ⫽ 568.011 The standard error of estimate of the cooling capacity for a specific living area is 568.0 Btu per hour. 24. (a) r 2 ⬇ 0.446 44.6% of the variation in y is explained. 55.4% of the variation in y is unexplained. (b) se ⫽ 235.079 The standard error of the price for a specific area is $235.08. 25. y ⫽ 0.757共64兲 ⫹ 21.525 ⫽ 69.973 ^
冪1 ⫹ n1 ⫹ n兺xn共x ⫺⫺ 共x兺x兲 兲 2
E ⫽ tc se
2
2
64 ⫺ 64.727兲 冪1 ⫹ 111 ⫹ 1111共共46,154 兲 ⫺ 712
2
⬇ 共1.833兲共2.206兲
2
⫽ 共1.833兲共2.202兲冪1.09866 ⬇ 4.231 y⫺E < y < y⫹E 69.973 ⫺ 4.231 < y < 69.973 ⫹ 4.231 65.742 < y < 74.204 ^
^
You can be 90% confident that the height of a male will be between 65.742 inches and 74.204 inches when his sister is 64 inches tall. 26. y ⫽ 193.845 ⫹ 3.431共74兲 ⫽ 447.739 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x兲2 ⫹ n n兺x2 ⫺ 共兺x)2 9关74 ⫺ 共716.9兾9兲兴 冪1 ⫹ 19 ⫹ 9共57,367.95 兲 ⫺ 共716.9兲 2
⫽ 共1.895兲共5.278兲
2
⫽ 共1.895兲共5.278兲冪1.233 ⫽ 11.105
y⫺E < y < y⫹E ^
^
447.739 ⫺ 11.105 < y < 447.739 ⫹ 11.105 436.634 < y < 458.844 You can be 90% confident that the average time spent watching television per household will be between 436.634 minutes and 458.844 minutes when 74 million households have multiple television sets. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 9

CORRELATION AND REGRESSION
27. y ⫽ ⫺0.086共45兲 ⫹ 10.450 ⫽ 6.580 ^
冪1 ⫹ n1 ⫹ n兺xn共x ⫺⫺ 共x兺x兲 兲 2
E ⫽ tc se
2
2
关45 ⫺ 共337兾7兲兴 冪1 ⫹ 17 ⫹ 77共18,563 兲 ⫺ 共337兲 2
⬇ 共2.571兲共0.623兲
2
⫽ 共2.571兲共0.623兲冪1.14708 ⫽ 1.715 y⫺E < y < y⫹E 6.580 ⫺ 1.715 < y < 6.58 ⫹ 1.715 4.865 < y < 8.295 ^
^
You can be 95% confident that the hours slept will be between 4.865 and 8.295 hours for a person who is 45 years old. 28. y ⫽ ⫺0.090共265兲 ⫹ 44.675 ⫽ 20.825 ^
n共x ⫺ x兲2
冪1 ⫹ n ⫹ n兺x ⫺ 共兺x兲 1 7共265 ⫺ 216.6兲 ⫽ 共2.571兲共1.476兲冪1 ⫹ ⫹ 7 7共369,382兲 ⫺ 1516 1
E ⫽ tc se
2
2
2
2
⫽ 共2.571兲共1.476兲 冪1.20133 ⬇ 4.159 y⫺E < y < y⫹E 20.825 ⫺ 4.159 < y < 20.825 ⫹ 4.159 16.668 < y < 24.982 ^
^
You can be 95% confident that the fuel efficiency will be between 16.668 miles per gallon and 24.982 miles per gallon when the engine displacement is 265 cubic inches. 29. y ⫽ 3002.991 ⫹ 9.468共720兲 ⫽ 9819.95 1 n共x ⫺ x兲2 E ⫽ tc se 1⫹ ⫹ ⫽ 共3.707兲共568兲 n n兺x2 ⫺ 共兺x兲2 ^
冪
8共720 ⫺ 499.4兲 冪1 ⫹ 18 ⫹ 8共2,182,275 兲 ⫺ 3995 2
2
⫽ 共3.707兲共568兲冪1.38486 ⬇ 2477.94 y⫺E < y < y⫹E 9819.95 ⫺ 2477.94 < y < 9819.95 ⫹ 2477.94 7342.01 < y < 12,297.89 ^
^
You can be 99% confident that the cooling capacity will be between 7342.01 Btu per hour and 12,297.89 Btu per hour when the living area is 720 square feet. 30. y ⫽ ⫺532.053 ⫹ 1.454x ⫽ ⫺532.053 ⫹ 1.454共900兲 ⫽ 776.547 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x兲2 ⫹ n n兺x2 ⫺ 共兺x兲2
共900 ⫺ 733.5兲 冪1 ⫹ 181 ⫹ 18共182,314,270 兲 ⫺ 6012 2
⫽ 2.921共235.079兲
2
⫽ 共2.921兲共235.079兲 冪1.13789 ⬇ 732.480
y⫺E < y < y⫹E 776.547 ⫺ 732.480 < y < 776.547 ⫹ 732.480 44.067 < y < 1509.027 ^
^
You can be 99% confident that the price will be between $44.07 and $1509.03 when the cooking area is 900 square inches. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
283
284
CHAPTER 9

CORRELATION AND REGRESSION
31. y ⫽ 3.674 ⫹ 1.287x1 ⫹ ⫺7.531x2 ^
32. se ⫽ 0.710; r 2 ⫽ 0.943 94.3% of the variation in y can be explained by the model. 33. (a) 21.705
(b) 25.21
(c) 30.1
(d) 25.86
34. (a) 11.272
(b) 14.695
(c) 14.380
(d) 9.232
CHAPTER 9 QUIZ SOLUTIONS y
Personal outlays (in trillions of dollars)
1.
The data appear to have a positive linear correlation. The outlays increase as the incomes increase.
10 9 8 7 6 5 x 6 7 8 9 10 11
Personal income (in trillions of dollars)
2. r ⬇ 0.996 → Strong positive linear correlation 3. H0 : ⫽ 0; Ha: ⫽ 0
␣ ⫽ 0.05, d.f. ⫽ n ⫺ 2 ⫽ 9 t0 ⫽ ± 2.262 t⫽
r
⫽
0.996
兲 冪1n ⫺⫺ r2 冪1 ⫺11共0.996 ⫺2 2
2
⫽
0.996 ⬇ 33.44 冪0.00088711
Reject H0 . There is enough evidence to conclude that a significant correlation exists. y
Personal outlays (in trillions of dollars)
4. 兺y2 ⫽ 616.090 兺y ⫽ 81.100 兺xy ⫽ 715.940 b ⫽ ⫺1.018 m ⫽ 0.976 y ⫽ 0.976x ⫺ 1.018 ^
10 9 8 7 6 5 x 6 7 8 9 10 11
Personal income (in trillions of dollars)
5. y ⫽ 0.976共6.2兲 ⫺ 1.018 ⫽ $5.033 trillion ^
6. r 2 ⫽ 0.993 99.3% of the variation in y is explained. 0.7% of the variation in y is unexplained. 7. se ⫽ 0.121 The standard deviation of personal outlays for a specified personal income is $0.121 trillion.
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CHAPTER 9

CORRELATION AND REGRESSION
8. y ⫽ 0.976共7.6兲 ⫺ 1.018 ⫽ 6.400 ^
冪
E ⫽ tc se
1⫹
1 n共x ⫺ x 兲2 ⫹ n n兺x2 ⫺ 共兺x兲2 ⫺ 共94.6兾11兲兴 冪1 ⫹ 111 ⫹ 1111关7.6共832.5 兲 ⫺ 94.6
2
⫽ 2.262共0.120兲
2
⬇ 共2.262兲共0.120兲冪1.144 ⬇ 0.290 y⫺E < y < y⫹E 6.400 ⫺ 0.290 < y < 6.400 ⫹ 0.290 6.110 < y < 6.690 → ($6.110 trillion, $6.690 trillion) ^
^
You can be 95% confident that the personal outlays will be between $6.110 trillion and $6.690 trillion when personal income is $7.6 trillion. 9. (a) 1374.762 pounds (b) 1556.755 pounds (c) 1183.262 pounds (d) 1294.385 x2 has the greater influence on y.
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285
ChiSquare Tests and the FDistribution
CHAPTER
10
10.1 GOODNESS OF FIT
10.1 Try It Yourself Solutions 1a.
Music Classical Country Gospel Oldies Pop Rock
% of Listeners
Expected Frequency
4% 36% 11% 2% 18% 29%
12 108 33 6 54 87
2a. The expected frequencies are 64, 80, 32, 56, 60, 48, 40, and 20, all of which are at least 5. b. Claimed Distribution: Ages
Distribution
0–9 10–19 20–29 30–39 40–49 50–59 60–69 70+
16% 20% 8% 14% 15% 12% 10% 5%
H0 : Distribution of ages is as shown in table above. Ha: Distribution of ages differs from the claimed distribution. c. 0.05
d. d.f. n 1 7
e. 20 14.067; Reject H0 if 2 > 14.067. f. Ages
Distribution
Observed
Expected
0–9 10–19 20–29 30–39 40–49 50–59 60–69 70+
16% 20% 8% 14% 15% 12% 10% 5%
76 84 30 60 54 40 42 14
64 80 32 56 60 48 40 20
(O ⴚ E )2 E 2.250 0.200 0.125 0.286 0.600 1.333 0.100 1.800 6.694
2 ⬇ 6.694 g. Fail to reject H0 . h. There is not enough evidence to suppport the claim that the distribution of ages differs from the claimed distribution.
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288
CHAPTER 10

CHISQUARE TESTS AND THE FDISTRIBUTION
3a. The expected frequencies are 123, 138, and 39, all of which are at least 5. b. Claimed Distribution: Response
Distribution
Retirement Children’s college education Not sure
41% 46% 13%
H0 : Distribution of responses is as shown in table above. (claim) Ha: Distribution of responses differs from the claimed distribution. c. 0.01
e. 20 9.210; Reject H0 if 2 > 9.210.
d. d.f. n 1 2
f. Response
Distribution
Observed
Expected
(O ⴚ E )2 E
41% 46% 13%
129 149 22
123 138 39
0.293 0.877 7.410
Retirement Children’s college education Not sure
8.580
2 ⬇ 8.580 g. Fail to reject H0 . h. There is not enough evidence to dispute the claimed distribution of responses. 4a. The expected frequencies are 30 for each color. b. H0 : The distribution of differentcolored candies in bags of peanut M&M’s is uniform. (claim) H1: The distribution of differentcolored candies in bags of peanut M&M’s is not uniform. c. 0.05
d. d.f. k 1 6 1 5
f. Color
Observed
Expected
(O ⴚ E )2 E
22 27 22 41 41 27
30 30 30 30 30 30
2.133 0.300 2.133 4.033 4.033 0.300
Brown Yellow Red Blue Orange Green
2
兺
e. 20 11.071; Reject H0 if 2 > 11.071.
共O E 兲 2 12.933 E
g. Reject H0 . h. There is enough evidence to dispute the claimed distribution.
10.1 EXERCISE SOLUTIONS 1. A multinomial experiment is a probability experiment consisting of a fixed number of trials in which there are more than two possible outcomes for each independent trial. 2. The observed frequencies must be obtained using a random sample, and each expected frequency must be greater than or equal to 5.
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CHAPTER 10

CHISQUARE TESTS AND THE FDISTRIBUTION
3. Ei npi 共150兲共0.3兲 45
4. Ei npi 共500兲共0.9兲 450
5. Ei npi 共230兲共0.25兲 57.5
6. Ei npi 共415兲共0.08兲 33.2
7. (a) Claimed distribution: Response
Distribution
Reward Program Low Interest Rate Cash Back Special Store Discounts Other
28% 24% 22% 8% 16%
H0: Distribution of responses is as shown in the table above. Ha: Distribution of responses differs from the claimed or expected distribution. (claim) (b) 20 9.488; Reject H0 if 2 > 9.488. (c) Response
Distribution
Observed
28% 24% 22% 8% 18%
112 98 107 46 62
Reward Program Low Interest Rate Cash Back Special Store Discounts Other
Expected
(O ⴚ E )2 E
119 102 93.5 34 76.5
0.412 0.157 1.949 4.235 2.748 9.501
2 9.501 (d) Reject H0. There is enough evidence at the 5% significance level to conclude that the distribution of responses differs from the claimed or expected distribution. 8. (a) Claimed distribution: Response Home Work Commuting Other
Distribution 70% 17% 8% 5%
H0 : Distribution of responses is as shown in table above. Ha: Distribution of responses differs from the claimed or expected distribution. (claim) (b) 20 7.815; Reject H0 if 2 > 7.815. (c) Response Home Work Commuting Other
Distribution
Observed
Expected
(O ⴚ E )2 E
70% 17% 8% 5%
389 110 55 27
406.70 98.77 46.48 29.05
0.770 1.277 1.562 0.145 3.754
2 ⬇ 3.754 (d) Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that the distribution of the responses differs from the claimed or expected distribution.
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9. (a) Claimed distribution: Response
Distribution
Keeping Medicare, Social Security and Medicaid Tax cuts Not sure
77% 15% 8%
H0: Distribution of responses is as shown in the table above. Ha: Distribution of responses differs from the claimed or expected distribution. (claim) (b) 20 9.210; Reject H0 if 2 > 9.210. (c) Response
Distribution
Observed
Expected
(O ⴚ E )2 E
77%
431
462
2.080
15% 8%
107 62
90 48
3.211 4.083
Keeping Medicare, Social Security, and Medicaid Tax cuts Not sure
9.374
2 9.374 (d) Reject H0. There is enough evidence at the 1% significance level to conclude that the distribution of responses differs from the claimed or expected distribution. 10. (a) Claimed distribution: Response Limited advancement Lack of recognition Low salary Unhappy with mgmt. Bored/don’t know
Distribution 41% 25% 15% 10% 9%
H0: Distribution of responses is as shown in table above. Ha: Distribution of responses differs from the claimed or expected distribution. (claim) (b) 20 13.277; Reject H0 if 2 > 13.277. (c) Response Limited advancement Lack of recognition Low salary Unhappy with mgmt. Bored/don’t know
Distribution
Observed
Expected
(O ⴚ E )2 E
41% 25% 15% 10% 9%
78 52 30 25 15
82.00 50.00 30.00 20.00 18.00
0.195 0.080 0.000 1.250 0.500 2.025
2 ⬇ 2.025 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to conclude that the distribution of the responses differs from the claimed or expected distribution.
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11. (a) Claimed distribution: Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Distribution 14.286% 14.286% 14.286% 14.286% 14.286% 14.286% 14.286%
H0 : The distribution of fatal bicycle accidents throughout the week is uniform as shown in the table above. (claim) Ha: The distribution of fatal bicycle accidents throughout the week is not uniform. (b) 20 10.645; Reject H0 if 2 > 10.645. (c) Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Distribution
Observed
Expected
(O ⴚ E )2 E
14.286% 14.286% 14.286% 14.286% 14.286% 14.286% 14.286%
108 112 105 111 123 105 118
111.717 111.717 111.717 111.717 111.717 111.717 111.717
0.124 0.0007 0.404 0.005 1.140 0.404 0.353 2.430
2 ⬇2.430 (d) Fail to reject H0 . There is not enough evidence at the 10% significance level to reject the claim that the distribution of fatal bicycle accidents throughout the week is uniform. 12. (a) Claimed distribution: Month January February March April May June July August September October November December
Distribution 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333%
H0: The distribution of fatal bicycle accidents is uniform by month as shown in table above. Ha: The distribution of fatal bicycle accidents is not uniform by month. (b) 20 17.275; Reject H0 if 2 > 17.275.
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(c) Month January February March April May June July August September October November December
Distribution
Observed
Expected
(O ⴚ E )2 E
8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333% 8.333%
37 42 43 62 79 93 66 86 81 94 57 43
65.247 65.247 65.247 65.247 65.247 65.247 65.247 65.247 65.247 65.247 65.247 65.247
12.229 8.283 7.585 0.162 2.899 11.805 0.009 6.6001 3.803 12.671 1.042 7.585 74.673
2 74.673 (d) Reject H0. There is enough evidence at the 10% significance level to conclude that the distribution of fatal bicycle accidents is not uniform by month. 13. (a) Claimed distribution: Object struck
Distribution
Tree Utility pole Embankment Guardrail Ditch Culvert Other
50% 14% 6% 5% 3% 2% 20%
H0: Distribution of objects struck is as shown in table above. Ha: Distribution of objects struck differs from the claimed or expected distribution. (claim) (b) 20 16.812; Reject H0 if 2 > 16.812. (c) Object Struck
Distribution
Observed
Expected
(O ⴚ E )2 E
Tree Utility pole Embankment Guardrail Ditch Culvert Other
50% 14% 6% 5% 3% 2% 20%
375 86 53 42 27 18 90
345.50 96.74 41.46 34.55 20.73 13.82 138.2
2.519 1.192 3.212 1.606 1.896 1.264 16.811
691
28.501
2 ⬇ 28.501 (d) Reject H0. There is enough evidence at the 1% significance level to conclude that the distribution of objects struck changed from the claimed or expected distribution.
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14. (a) Claimed distribution: Time of day Midnight–6 A.M. 6 A.M.–Noon Noon– 6 P.M. 6 P.M.–Midnight
Distribution 32% 15% 24% 29%
H0: Distribution of the time of day of roadside hazard crash deaths is as shown in table above. Ha: Distribution of the time of day of roadside hazard crash deaths differs from the claimed or expected distribution. (claim) (b) 20 11.345; Reject H0 if 2 > 11.345. (c) Time of day Midnight–6 A.M. 6 A.M.–Noon Noon–6 P.M. 6 P.M.–Midnight
Distribution
Observed
Expected
(O ⴚ E )2 E
32% 15% 24% 29%
224 128 115 160
200.64 94.05 150.48 181.83
2.720 12.255 8.365 2.621 25.961
2 25.961 (d) Reject H0. There is enough evidence at the 1% significance level to conclude that the distribution of the time of day of roadside hazard crash deaths has changed from the claimed or expected distribution. 15. (a) Claimed distribution: Response Not a HS grad HS graduate College (1yr+)
Distribution 33.333% 33.333% 33.333%
H0: Distribution of the responses is uniform as shown in table above. (claim) Ha: Distribution of the responses is not uniform. (b) 20 7.378; Reject H0 if 2 > 7.378. (c) Response Not a HS grad HS graduate College (1yr+)
Distribution
Observed
Expected
(O ⴚ E )2 E
33.333% 33.333% 33.333%
37 40 22
33 33 33
0.485 1.485 3.667
99
5.637
2 ⬇ 5.637 (d) Fail to reject H0. There is not enough evidence at the 2.5% significance level to reject the claim that the distribution of the responses is uniform.
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16. (a) Claimed distribution: Season
Distribution
Spring Summer Fall Winter
25% 25% 25% 25%
H0: Distribution of births is uniform as shown in table above. (claim) Ha: Distribution of births is not uniform. (b) 20 11.345; Reject H0 if 2 > 11.345. (c) Season
Distribution
Observed
Expected
(O ⴚ E )2 E
25% 25% 25% 25%
564 603 555 525
561.75 561.75 561.75 561.75
0.009 3.029 0.081 2.404
Spring Summer Fall Winter
2247
5.523
2 5.523 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to reject the doctor’s claim. 17. (a) Claimed distribution: Cause Trans. Accidents Objects/equipment Assaults Falls Exposure Fire/explosions
Distribution 43% 18% 14% 13% 9% 3%
H0 : Distribution of the causes is as shown in table above. Ha: Distribution of the causes differs from the claimed or expected distribution. (claim) (b) 20 11.071; Reject H0 if 2 > 11.071. (c) Cause Trans. Accidents Objects/equipment Assaults Falls Exposure Fires/explosions
Distribution
Observed
Expected
(O ⴚ E )2 E
43% 18% 14% 13% 9% 3%
2891 1159 804 754 531 92
2679.33 1121.58 872.34 810.03 560.79 186.93
16.722 1.248 5.354 3.876 1.582 48.209
6231
76.992
2 ⬇ 76.992 (d) Reject H0. There is enough evidence at the 5% significance level to conclude that the distributions of the causes in the Western U.S. differs from the national distribution.
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18. (a) Claimed distribution: Response
Distribution
Did not work Work 1–20 hours Work 21–34 hours Work 35+ hours
22% 26% 18% 34%
H0: Distribution of the responses is as shown in the table above. (claim) Ha: Distribution of the responses differs from the claimed or expected distribution. (b) 20 11.345; Reject H0 if 2 > 11.345. (c) Response
Distribution
Observed
Expected
(O ⴚ E )2 E
22% 26% 18% 34%
29 26 25 40
26.4 31.2 21.6 40.8
0.256 0.867 0.535 0.016
Did not work Work 1–20 hours Work 25–34 hours Work 35+ hours
120
1.674
2 1.674 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to reject the council’s claim. 19. (a) Frequency distribution: 69.435; ⬇ 8.337 Lower Boundary
Upper Boundary
Lower zscore
Upper zscore
Area
49.5 58.5 67.5 76.5 85.5
58.5 67.5 76.5 85.5 94.5
2.39 1.31 0.23 0.85 1.93
1.31 0.23 0.85 1.93 3.01
0.0867 0.3139 0.3933 0.1709 0.0255
Class Boundaries
Distribution
Frequency
Expected
(O ⴚ E ) 2 E
49.5–58.5 58.5–67.5 67.5–76.5 76.5–85.5 85.5–94.5
8.67% 31.39% 39.33% 17.09% 2.55%
19 61 82 34 4
17 63 79 34 5
0.235 0.063 0.114 0 0.2
200
0.612
H0: Test scores have a normal distribution. (claim) Ha: Test scores do not have a normal distribution. (b) 20 13.277; Reject H0 if 2 > 13.277. (c) 2 0.612 (d) Fail to reject H0 . There is not enough evidence at the 1% significance level to reject the claim that the distribution of test scores is normal.
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20. (a) Frequency distribution: 74.775, 9.822 Lower Boundary
Upper Boundary
Lower zscore
Upper zscore
Area
50.5 60.5 70.5 80.5 90.5
60.5 70.5 80.5 90.5 100.5
2.47 1.45 0.44 0.58 1.60
1.45 0.44 0.58 1.60 2.62
0.0668 0.2564 0.3891 0.2262 0.0504
Class Boundaries
Distribution
Frequency
Expected
(O ⴚ E ) 2 E
50.5–60.5 60.5–70.5 70.5–80.5 80.5–90.5 90.5–100.5
6.68% 25.64% 38.91% 22.62% 5.04%
28 106 151 97 18
27 103 156 90 20
0.037 0.087 0.160 0.544 0.2
400
1.028
H0: Test scores have a normal distribution. Ha: Test scores do not have a normal distribution. (b) 20 9.488; Reject H0 if 2 > 9.488. (c) 2 ⬇ 1.028 (d) Fail to reject H0. There is not enough evidence at the 5% significance level to reject the claim that the distribution of test scores is normal.
10.2 INDEPENDENCE
10.2 Try It Yourself Solutions 1ab.
Hotel
Leg Room
Rental Size
Other
Total
Business Leisure
36 38
108 54
14 14
22 14
180 120
Total
74
162
28
36
300
Hotel
Leg Room
Rental Size
Other
44.4 29.6
97.2 64.8
16.8 11.2
21.6 14.4
c. n 300 d. Business Leisure
2a. H0 : Travel concern is independent of travel purpose. Ha: Travel concern is dependent on travel purpose. (claim) b. 0.01 e. 2
兺
c. 共r 1兲共c 1兲 3
d. 20 11.345; Reject H0 if 2 > 11.345.
共O E兲2 共36 44.4兲2 共108 97.2兲2 共14 14.4兲2 . . . ⬇ 8.158 E 44.4 97.2 14.4
f. Fail to reject H0 . g. There is not enough evidence to conclude that travel concern is dependent on travel purpose. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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3a. H0: The number of minutes adults spend online per day is independent on gender. Ha: The number of minutes adults spend online per day is dependent on gender. (claim) b. Enter the data.
c. 20 9.488; Reject H0 if 2 > 9.488.
d. 2 ⬇ 65.619
e. Reject H0.
f. There is enough evidence to conclude that minutes spent online per day is dependent on gender.
10.2 EXERCISE SOLUTIONS 1. Er, c
共Sum of row r兲共Sum of column c兲 Sample size
2. A large chisquare test statistic is evidence for rejecting the null hypothesis. So, the chisquare independence test is always a righttailed test. 3. False. In order to use the 2 independence test, each expected frequency must be greater than or equal to 5. 4. False. A contingency table with two rows and five columns will have 共2 1兲共5 1兲 4 degrees of freedom. 5. False. If the two variables of a chisquare test for independence are dependent, then you can expect to find large differences between the observed frequencies and the expected frequencies. 6. True 7. (a)
Athlete has Result Injury No injury
(b)
Stretched
Not stretched
18 211
22 189
40 400
229
211
440
Athlete has Result Injury No injury
Stretched
Not stretched
20.818 208.182
19.182 191.818
8. (a)
Treatment Result Got the flu Did not get the flu
(b)
Flu shot
No flu shot
54 196
157 103
211 299
250
260
510
Treatment Result
Flu shot
No flu shot
Got the flu Did not get the flu
103.431 146.569
107.569 152.431
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9. (a)
Treatment Result
Brandname
Generic
Placebo
24 12
21 13
10 45
55 70
36
34
55
125
Brandname
Generic
Placebo
15.84 20.16
14.96 19.04
24.2 30.8
Improvement No change
(b)
Treatment Result Improvement No change
10. (a)
Rating Size
Excellent
Fair
Poor
182 180
203 311
165 159
550 650
362
514
324
1200
Seats 100 or fewer Seats over 100
(b)
Rating Size
Excellent
Seats 100 or fewer Seats over 100
Fair
165.917 196.083
11. (a)
Poor
235.583 278.417
148.500 175.500
Type of Car Gender
Compact
Fullsize
SUV
Truck/Van
Male Female
28 24
39 32
21 20
22 14
110 90
52
71
41
36
200
(b)
Type of Car Gender
Compact
Fullsize
SUV
Truck/Van
Male Female
28.60 23.40
39.05 31.95
22.55 18.45
19.80 16.20
12. (a)
Age Result Willing to buy hybrid Not willing to buy hybrid
(b)
18–30
31–42
43–61
62+
72 14
66 21
73 19
69 21
280 75
86
87
92
90
355
Age Result
18–30
31–42
43–61
62+
Willing to buy hybrid Not willing to buy hybrid
67.831 18.169
68.620 18.380
72.563 19.437
70.986 19.014
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13. (a) H0: Skill level in a subject is independent of location. (claim) Ha: Skill level in a subject is dependent on location. (b) d.f. 共r 1兲共c 1兲 2
20 9.210; Reject H0 if 2 > 9.210. (c) 2 ⬇ 0.297 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to reject the claim that skill level in a subject is independent of location. 14. (a) H0: Attitudes about safety are independent of the type of school. Ha: Attitudes about safety are dependent on the type of school. (claim) (b) d.f. 共r 1兲共c 1兲 1
20 6.635; Reject H0 if 2 > 6.635. (c) 2 ⬇ 8.691 (d) Reject H0. There is enough evidence at the 1% significance level to conclude that attitudes about safety are dependent on the type of school. 15. (a) H0: Grades are independent of the institution. Ha: Grades are dependent on the institution. (claim) (b) d.f. 共r 1兲共c 1兲 8
20 15.507; Reject H0 if 2 > 15.507. (c) 2 ⬇ 48.488 (d) Reject H0. There is enough evidence at the 5% significance level to conclude that grades are dependent on the institution. 16. (a) H0: Adults’ ratings are independent of the type of school. Ha: Adults’ ratings are dependent on the type of school. (claim) (b) d.f. 共r 1兲共c 1兲 4
20 7.815; Reject H0 if 2 > 7.815. (c) 2 ⬇ 148.389 (d) Reject H0. There is enough evidence at the 5% significance level to conclude that adults’ ratings are dependent on the type of school. 17. (a) H0: Results are independent of the type of treatment. Ha: Results are dependent on the type of treatment. (claim) (b) d.f. 共r 1兲共c 1兲 1
20 2.706; Reject H0 if 2 > 2.706. (c) 2 ⬇ 5.106 (d) Reject H0. There is enough evidence at the 10% significance level to conclude that results are dependent on the type of treatment. I would recommend using the drug.
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18. (a) H0: Results are independent of the type of treatment. Ha: Results are dependent on the type of treatment. (claim) (b) d.f. 共r 1兲共c 1兲 1
20 2.706; Reject H0 if 2 > 2.706. (c) 2 ⬇ 1.032 (d) Fail to reject H0. There is not enough evidence at the 10% significance level to conclude that results are dependent on the type of treatment. I would not recommend using the drug. 19. (a) H0: Reasons are independent of the type of worker. Ha: Reasons are dependent on the type of worker. (claim) (b) d.f. 共r 1兲共c 1兲 2
20 9.210; Reject H0 if 2 > 9.210. (c) 2 ⬇ 7.326 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to conclude that the reason(s) for continuing education are dependent on the type of worker. Based on these results, marketing strategies should not differ between technical and nontechnical audiences in regard to reason(s) for continuing education. 20. (a) H0: Most important aspect of career development is independent of age. Ha: Most important aspect of career development is dependent on age. (claim) (b) d.f. 共r 1兲共c 1兲 4
20 13.277; Reject H0 if 2 > 13.277. (c) 2 ⬇ 5.757 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to conclude that the most important aspect of career development is dependent on age. 21. (a) H0: Type of crash is independent of the type of vehicle. Ha: Type of crash is dependent on the type of vehicle. (claim) (b) d.f. 共r 1兲共c 1兲 2
20 5.991; Reject H0 if 2 > 5.991. (c) 2 ⬇ 108.913 (d) Reject H0. There is enough evidence at the 5% significance level to conclude that type of crash is dependent on the type of vehicle. 22. (a) H0: Age is independent of gender. Ha: Age is dependent on gender. (claim) (b) d.f. 共r 1兲共c 1兲 5
20 11.071; Reject H0 if 2 > 11.071. (c) 2 ⬇ 1.997 (d) Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that age is dependent on gender in such alcoholrelated accidents.
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23. (a) H0: Subject is independent of coauthorship. Ha: Subject and coauthorship are dependent. (claim) (b) d.f. 共r 1兲共c 1兲 4
20 7.779; Reject H0 if 2 > 7.779. (c) 2 ⬇ 5.610 (d) Fail to reject H0. There is not enough evidence at the 10% significance level to conclude that the subject matter and coauthorship are related. 24. (a) H0: Access speed is independent of metropolitan status. H1: Access speed is dependent on metropolitan status. (claim) (b) d.f. 共r 1兲共c 1兲 4
20 13.277; Reject H0 if 2 > 13.277. (c) 2 18.145 (d) Reject H0. There is not enough evidence at the 1% significance level to conclude that access speed is dependent on metropolitan status. 25. H0: The proportions are equal. (claim) Ha: At least one of the proportions is different from the others. d.f. 共r 1兲共c 1兲 7
20 14.067; Reject H0 if 2 > 14.067. 2 ⬇ 7.462 Fail to reject H0. There is not enough evidence at the 5% significance level to reject the claim that the proportions are equal. 26. (a) H0: The proportions are equal. (claim) H1: At least one proportion is different from the others. (b) d.f. 共r 1兲共c 1兲 1
20 2.706; Reject H0 if 2 > 2.706. (c) 2 5.106 (d) Reject H0. There is not enough evidence at the 10% significance level to reject the claim that the proportions are equal. 27.
Educational Attainment Not a high school graduate
High school graduate
Some college, no degree
Employed
0.0610
0.1903
0.1129
0.2725
Unemployed
0.0058
0.0111
0.0053
0.0074
Not in labor force
0.0817
0.1204
0.0498
0.0817
Status
28. 11.3%
29. 8.2%
31. 63.7%
32. 14.9%
Associate’s, Bachelor’s, or advanced degree
30. 6.1%
33. Several of the expected frequencies are less than 5.
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34.
Educational Attainment Not a high school graduate
High school graduate
Some college, no degree
Employed
0.0958
0.2989
0.1774
0.4280
Unemployed
0.1964
0.3750
0.1786
0.2500
Not in labor force
0.2448
0.3609
0.1494
0.2448
Status
35. 29.9%
Associate’s, Bachelor’s, or advanced degree
36. 14.9%
37.
Educational Attainment Not a high school graduate
High school graduate
Some college, no degree
Employed
0.4107
0.5914
0.6719
0.7537
Unemployed
0.0393
0.0346
0.0315
0.0205
Not in labor force
0.5500
0.3740
0.2965
0.2258
Status
38. 22.6%
Percent
40.
Associate’s, Bachelor’s, or advanced degree
39. 3.9%
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 N H S A N H S A N H S A
Employed Unemployed
Not in labor force
41. As educational attainment increases, employment increases.
10.3 COMPARING TWO VARIANCES
10.3 Try It Yourself Solutions 1a. 0.01
b. F 5.42
2a. 0.01
b. F 18.31
3a. H0 : 21 22 ; Ha : 21 > 22 (claim) b. 0.01 c. d.f.N n1 1 24 d.f.D n2 1 19 d. F0 2.92; Reject H0 if F > 2.92. e. F
s 21 180 ⬇ 3.21 s 22 56
f. Reject H0 . g. There is enough evidence to support the claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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4a. H0 : 1 2 (claim); Ha: 1 2 b. 0.01 c. d.f.N n1 1 15 d.f.D n2 1 21 d. F0 3.43; Reject H0 if F > 3.43. e. F
s 21 共0.95兲2 ⬇ 1.48 s 22 共0.78兲2
f. Fail to reject H0. g. There is not enough evidence to reject the claim.
10.3 EXERCISE SOLUTIONS 1. Specify the level of significance . Determine the degrees of freedom for the numerator and denominator. Use Table 7 in Appendix B to find the critical value F. 2. (1) The Fdistribution is a family of curves determined by two types of degrees of freedom, d.f.N and d.f.D. (2) Fdistributions are positively skewed. (3) The area under the Fdistribution curve is equal to 1. (4) Fvalues are always greater than or equal to zero. (5) For all Fdistributions, the mean value of F is approximately equal to 1. 3. (1) The samples must be randomly selected, (2) The samples must be independent, and (3) Each population must have a normal distribution. 4. Determine the sample whose variance is greater. Use the size of this sample to find d.f.N. Use the size of the other sample to find d.f.D. 5. F 2.54
6. F 5.61
7. F 4.86
8. F 2.63
9. F 2.06
10. F 14.62
11. H0 : 21 22 ; Ha : 21 > 22 (claim) d.f.N 4 d.f.D 5 F0 3.52; Reject H0 if F > 3.52. F
s21 773 ⬇ 1.010 s 22 765
Fail to reject H0. There is not enough evidence to support the claim.
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12. H0: 21 22 (claim); Ha: 21 22 d.f.N 6 d.f.D 7 F0 5.12; Reject H0 if F > 5.12. F
s 21 310 ⬇ 1.044 s 22 297
Fail to reject H0. There is not enough evidence to reject the claim. 13. H0 : 21 22 (claim); Ha : 21 > 22 d.f.N 10 d.f.D 9 F0 5.26; Reject H0 if F > 5.26. F
s21 842 ⬇ 1.007 s 22 836
Fail to reject H0. There is not enough evidence to reject the claim. 14. H0: 21 22 ; Ha: 21 22 (claim) d.f.N 30 d.f.D 27 F0 1.88; Reject H0 if F > 1.88. F
s 21 245 ⬇ 2.188 s 22 112
Reject H0. There is enough evidence to support the claim. 15. H0: 21 22 (claim); Ha : 21 22 d.f.N 12 d.f.D 19 F0 3.30; Reject H0 if F > 3.30. F
s 21 9.8 ⬇ 3.920 s 22 2.5
Reject H0. There is enough evidence to reject the claim. 16. H0: 21 22 ; Ha: 21 > 22 (claim) d.f.N 15 d.f.D 11 F0 2.72; Reject H0 if F > 2.72. F
s 21 44.6 ⬇ 1.135 s 22 39.3
Fail to reject H0. There is not enough evidence to support the claim.
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17. Population 1: Company B Population 2: Company A (a) H0 : 21 22 ; Ha : 21 > 22 (claim) (b) d.f.N 24 d.f.D 19 F0 2.11; Reject H0 if F > 2.11. (c) F
s 21 2.8 ⬇ 1.08 s 22 2.6
(d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to support Company A’s claim that the variance of life of its appliances is less than the variance of life of Company B appliances. 18. (a) Population 1: Competitor Population 2: Auto Manufacturer H0: 12 22 ; Ha: 21 > 22 (claim) (b) d.f.N 20 d.f.D 18 F0 2.19; Reject H0 if F > 2.19. (c) F
s 21 22.5 ⬇ 2.01 s 22 16.2
(d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to conclude that the variance of the fuel consumption for the company’s hybrid vehicles is less than that of the competitor’s hybrid vehicles. 19. Population 1: District 1 Population 2: District 2 (a) H0: 21 22 (claim); Ha: 21 22 (b) d.f.N 11 d.f.D 13 F0 2.635; Reject H0 if F > 2.635. (c) F
s 21 共36.8兲2 ⬇ 1.282 s 22 共32.5兲2
(d) Fail to reject H0. (e) There is not enough evidence at the 10% significance level to reject the claim that the standard deviation of science assessment test scores for eighth grade students is the same in Districts 1 and 2.
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20. (a) Population 1: District 1 Population 2: District 2 H0: 21 22 (claim); Ha: 21 22 (b) d.f.N 9 d.f.D 12 F0 5.20; Reject H0 if F > 5.20. (c) F
s 21 共33.9兲2 ⬇ 1.26 s 22 共30.2兲2
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to reject the claim that the standard deviation of U.S. history assessment test scores for eighth grade students is the same for Districts 1 and 2. 21. Population 1: Before new admissions procedure Population 2: After new admissions procedure (a) H0 : 21 22 ; Ha : 21 > 22 (claim) (b) d.f.N 24 d.f.D 20 F0 1.77; Reject H0 if F > 1.77. (c) F
s 21 共0.7兲2 ⬇ 1.96 s 22 共0.5兲2
(d) Reject H0. (e) There is enough evidence at the 10% significance level to support the claim that the standard deviation of waiting times has decreased. 22. (a) Population 1: 2nd city Population 2: 1st city H0: 12 22 (claim); Ha: 21 22 (b) d.f.N 30 d.f.D 27 F0 2.73; Reject H0 if F > 2.73. (c) F
s 21 共39.50兲2 ⬇ 2.77 s 22 共23.75兲2
(d) Reject H0. (e) There is enough evidence at the 1% significance level to reject the claim that the standard deviations of hotel room rates for the two cities are the same.
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CHAPTER 10

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23. (a) Population 1: New York Population 2: California H0 : 21 22 ; Ha : 21 > 22 (claim) (b) d.f.N 15 d.f.D 16 F0 2.35; Reject H0 if F > 2.35. (c) F
s 21 共14,900兲2 ⬇ 2.41 s 22 共9,600兲2
(d) Reject H0. (e) There is enough evidence at the 5% significance level to conclude the standard deviation of annual salaries for actuaries is greater in New York than in California. 24. (a) Population 1: Florida Population 2: Louisiana H0: 21 22 ; Ha: 21 > 22 (claim) (b) d.f.N 27 d.f.D 23 F0 1.985; Reject H0 if F > 1.985. (c) F
s 21 共10,100兲2 ⬇ 2.490 s 22 共6,400兲2
(d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that the standard deviation of the annual salaries for public relations managers is greater in Florida than in Louisiana. 25. Righttailed: FR 14.73 Lefttailed: (1) d.f.N 3 and d.f.D 6 (2) F 6.60 (3) Critical value is
1 1 ⬇ 0.15. F 6.60
26. Righttailed: F 2.33 Lefttailed: (1) d.f.N 15 and d.f.D 20 (2) F 2.20 (3) Critical value is
1 1 ⬇ 0.45. F 2.20
27.
21 21 10.89 21 s 21 s 21 10.89 F < < F → 0.331 < < 3.33 → 0.375 < < 3.774 s 22 L 22 s 22 R 9.61 22 9.61 22
28.
2 s2 5.29 2 5.29 2 s 21 FL < 21 < 21 FR → 0.331 < 12 < 3.33 → 0.485 < 12 < 4.880 2 s2 2 s2 3.61 2 3.61 2
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10.4 ANALYSIS OF VARIANCE
10.4 Try It Yourself Solutions 1a. H0 : 1 2 3 4 Ha: At least one mean is different from the others. (claim) b. 0.05 c. d.f.N 3 d.f.D 14 d. F0 3.34; Reject H0 if F > 3.34. e. Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
549.8 608.0
3 14
183.3 43.4
F 4.22
F ⬇ 4.22 f. Reject H0 . g. There is enough evidence to conclude that at least one mean is different from the others. 2a. Enter the data. b. H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
0.584 4.360
3 30
0.195 0.145
F 1.34
F 1.34 → Pvalue 0.280 c. 0.280 > 0.05 d. Fail to reject H0. There is not enough evidence to conclude that at least one mean is different from the others.
10.4 EXERCISE SOLUTIONS 1. H0: 1 2 . . . k Ha: At least one of the means is different from the others. 2. Each sample must be randomly selected from a normal, or approximately normal, population. The samples must be independent of each other. Each population must have the same variance. 3. MSB measures the differences related to the treatment given to each sample. MSW measures the differences related to entries within the same sample.
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CHAPTER 10

CHISQUARE TESTS AND THE FDISTRIBUTION
4. H0A: There is no difference among the treatment means of Factor A. HaA: There is at least one difference among the treatment means of Factor A. H0B: There is no difference among the treatment means of Factor B. HaB: There is at least one difference among the treatment means of Factor B. H0AB: There is no interaction between Factor A and Factor B. HaAB: There is no interaction between Factor A and Factor B. 5. (a) H0: 1 2 3 Ha: At least one mean is different from the others. (claim) (b) d.f.N k 1 2 d.f.D N k 26 F0 3.37; Reject H0 if F > 3.37. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
0.518 6.629
2 26
0.259 0.255
F 1.017
F ⬇ 1.02 (d) Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that the mean costs per ounce are different. 6. (a) H0: 1 2 3 Ha: At least one mean is different from the others. (claim) (b) d.f.N k 1 2 d.f.D N k 14 F0 3.74; Reject H0 if F > 3.74. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
342.549 4113.333
2 14
171.275 293.810
F 0.583
F ⬇ 0.58 (d) Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that at least one of the mean battery prices is different from the others. 7. (a) H0: 1 2 3 (claim) Ha: At least one mean is different from the others. (b) d.f.N k 1 2 d.f.D N k 12 F0 2.8l; Reject H0 if F > 2.81.
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Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
302.6 1024.2
2 12
151.3 85.4
F 1.77
F ⬇ 1.77 (d) Fail to reject H0. There is not enough evidence at the 10% significance level to reject the claim that the mean prices are all the same for the three types of treatment. 8. (a) H0: 1 2 3 4 (claim) Ha: At least one mean is different from the others. (b) d.f.N k 1 3 d.f.D N k 23 F0 2.34; Reject H0 if F > 2.34. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
13,466 77,234
3 23
4488.667 3358
F 1.337
F ⬇ 1.34 (d) Fail to reject H0. There is not enough evidence at the 10% significance level to reject the claim that the mean annual amounts are the same in all regions. 9. (a) H0: 1 2 3 4 (claim) Ha: At least one mean is different from the others. (b) d.f.N k 1 3 d.f.D N k 29 F0 4.54; Reject H0 if F > 4.54. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
5.608 97.302
3 29
1.869 3.355
F 0.557
F ⬇ 0.56 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to reject the claim that the mean number of days patients spend in the hospital is the same for all four regions. 10. (a) H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) (b) d.f.N k 1 3 d.f.D N k 50 F0 2.205; Reject H0 if F > 2.205.
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CHAPTER 10
(c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
216.277 977.796
3 50
72.092 19.556

CHISQUARE TESTS AND THE FDISTRIBUTION
F 3.686
F ⬇ 3.686 (d) Reject H0. There is enough evidence at the 10% significance level to conclude that the mean square footage for at least one of the four regions is different from the others. 11. (a) H0: 1 2 3 4 (claim) Ha: At least one mean is different from the others. (b) d.f.N k 1 3 d.f.D N k 40 F0 2.23; Reject H0 if F > 2.23. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
15,095.256 73,015.903
3 43
5031.752 1825.398
F 2.757
F ⬇ 2.76 (d) Reject H0. There is enough evidence at the 10% significance level to reject the claim that the mean price is the same for all four cities. 12. (a) H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) (b) d.f.N k 1 3 d.f.D N k 16 F0 3.24; Reject H0 if F > 3.24. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
115,823,113.8 240,248,350
3 16
38,607,704.58 15,015,521.88
F 2.571
F ⬇ 2.57 (d) Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that the mean salary is different in at least one of the areas. 13. (a) H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) (b) d.f.N k 1 3 d.f.D N k 26 F0 2.98; Reject H0 if F > 2.98
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Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
2,220,266,803 1,188,146,984
3 16
740,088,934 45,697,960.9
F 16.195
F ⬇ 16.20 (d) Reject H0. There is enough evidence at the 10% significance level to conclude that at least one of the mean prices is different from the others. 14. (a) H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) (b) d.f.N k 1 3 d.f.D N k 35 F0 2.255; Reject H0 if F > 2.255. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
13,935.933 57,600.361
3 35
4645.311 1645.725
F 2.823
F ⬇ 2.823 (d) Reject H0. There is enough evidence at the 1% significance level to conclude that the mean energy consumption for at least one region is different from the others. 15. (a) H0: 1 2 3 4 (claim) Ha: At least one mean is different from the others. (b) d.f.N k 1 3 d.f.D N k 28 F0 4.57; Reject H0 if F > 4.57. (c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
771.25 15,674.75
3 28
257.083 559.813
F 0.459
F ⬇ 0.46 (d) Fail to reject H0. There is not enough evidence at the 1% significance level to reject the claim that the mean numbers of female students are equal for all grades. 16. (a) H0: 1 2 3 4 (claim) Ha: At least one mean is different from the others. (b) d.f.N k 1 3 d.f.D N k 30 F0 4.51; Reject H0 if F > 4.51.
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CHAPTER 10
(c) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
1,428,783.49 8,309,413.01
3 30
476,261.162 276,980.434

CHISQUARE TESTS AND THE FDISTRIBUTION
F 1.719
F ⬇ 1.72 (d) Fail to reject H0. There is not enough evidence at the 1% level to reject the claim that the mean amounts spent are equal for all regions. 17. H0: Advertising medium has no effect on mean ratings. Ha: Advertising medium has an effect on mean ratings. H0: Length of ad has no effect on mean ratings. Ha: Length of ad has an effect on mean ratings. H0: There is no interaction effect between advertising medium and length of ad on mean ratings. Ha: There is an interaction effect between advertising medium and length of ad on mean ratings. Source
d.f.
SS
MS
F
P
Ad medium Length of ad Interaction Error
1 1 1 16
1.25 0.45 0.45 34.80
1.25 0.45 0.45 2.17
0.57 0.21 0.21
0.459 0.655 0.655
Total
19
36.95
None of the null hypotheses can be rejected at the 10% significance level. 18. H0: Type of vehicle has no effect on the mean number of vehicles sold. Ha: Type of vehicle has an effect on the mean number of vehicles sold. H0: Gender has no effect on the mean number of vehicles sold. Ha: Gender has an effect on the mean number of vehicles sold. H0: There is no interaction effect between type of vehicle and gender on the mean number of vehicles sold. Ha: There is an interaction effect between type of vehicle and gender on the mean number of vehicles sold. Source
d.f.
SS
MS
F
P
Type of vehicle Gender Interaction Error
2 1 2 18
84.08 0.38 12.25 22.25
42.04 0.38 6.12 1.24
34.01 0.30 4.96
0.000 0.589 0.019
Total
23
118.96
There appears to be an interaction effect between type of vehicle and gender on the mean number of vehicles sold. Also, there appears that type of vehicle has an effect on the mean number of vehicles sold.
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19. H0: Age has no effect on mean GPA. Ha: Age has an effect on mean GPA. H0: Gender has no effect on mean GPA. Ha: Gender has an effect on mean GPA. H0: There is no interaction effect between age and gender on mean GPA. Ha: There is an interaction effect between age and gender on mean GPA. Source
d.f.
SS
MS
F
P
Age Gender Interaction Error
3 1 3 16
0.41 0.18 0.29 18.66
0.14 0.18 0.10 1.17
0.12 0.16 0.08
0.948 0.697 0.968
Total
23
19.55
None of the null hypotheses can be rejected at the 10% significance level. 20. H0: Technicians have no effect on mean repair time. Ha: Technicians have an effect on mean repair time. H0: Brand has no effect on mean repair time. Ha: Brand has an effect on mean repair time. H0: There is no interaction effect between technicians and brand on mean repair time. Ha: There is an interaction effect between technicians and brand on mean repair time. Source
d.f.
SS
MS
F
P
Technicians Brand Interaction Error
3 2 6 24
714 382 2007 3277
238 191 334 137
1.74 1.40 2.45
0.185 0.266 0.054
Total
35
6381
There appears to be an interaction effect between technicians and brand on the mean repair time at the 10% significance level. 21.
Mean
Size
Pop 1
17.82
13
Pop 2
13.50
13
Pop 3
13.12
14
Pop 4
12.91
14
SSw 977.796 兺共ni 1兲 N k 50 F0 2.205 → CVScheffé 2.205共4 1兲 6.615
共x 1 x 2 兲2 SSw 1 1 兺共ni 1兲 n1 n2
冤
冥
共x 1 x 3兲2 SSw 1 1 兺共ni 1兲 n1 n 3
冤
冥
⬇ 6.212 → No difference
⬇ 7.620 → Significant difference
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CHAPTER 10
共x 1 x 4兲2 SSw 1 1 兺共ni 1兲 n1 n4
冤
共x 2 x 3 兲2 SSw 1 1 兺共ni 1兲 n2 n3
冤
共x 2 x 4 兲2 SSw 1 1 兺共ni 1兲 n2 n4
冤
冥 冥 冥
共x 3 x4 兲2 SSw 1 1 兺共ni 1兲 n 3 n4
冤
22.

CHISQUARE TESTS AND THE FDISTRIBUTION
⬇ 8.330 → Significant difference
⬇ 0.049 → No difference
⬇ 0.121 → No difference
冥
Mean
Size
Pop 1
216.67
11
Pop 2
213.56
10
Pop 3
197.22
12
Pop 4
247.70
11
⬇ 0.016 → No difference
SSw 73,015.903 兺共ni 1兲 N k 40 F0 2.23 → CVScheffé 2.23共4 1兲 6.69
共x 1 x2兲2 ⬇ 0.03 → No difference SSw 1 1 兺共ni 1兲 n1 n2
冤
冥
共x 1 x 3兲2 ⬇ 1.19 → No difference SSw 1 1 兺共ni 1兲 n1 n3
冤
冥
共x 1 x 4兲2 ⬇ 2.90 → No difference SSw 1 1 兺共ni 1兲 n1 n4
冤
冥
共x 2 x 3兲2 ⬇ 0.80 → No difference SSw 1 1 兺共ni 1兲 n2 n3
冤
冥
共x2 x4兲2 ⬇ 3.34 → No difference SSw 1 1 兺共ni 1兲 n2 n4
冤
冥
共x 3 x 4兲2 ⬇ 8.01 → Significant difference SSw 1 1 兺共ni 1兲 n3 n4
冤
冥
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23.

CHISQUARE TESTS AND THE FDISTRIBUTION
Mean
Size
Pop 1
66,492
8
Pop 2
60,528
7
Pop 3
55,504
9
Pop 4
79,500
6
SSw ⬇ 1,188,146,984 兺共ni 1兲 N k 26 F0 2.98 → CVScheffé 2.98共4 1兲 8.94
共x 1 x 2 兲2 SSw 1 1 兺共ni 1兲 n1 n2
冤
冥
共x 1 x 3兲2 SSw 1 1 兺共ni 1兲 n1 n 3
冤
共x 1 x 4 兲2 SSw 1 1 兺共ni 1兲 n1 n4
冤
共x2 x3 兲2 SSw 1 1 兺共ni 1兲 n2 n3
冤
共x2 x4 兲2 SSw 1 1 兺共ni 1兲 n2 n4
冤 冤
24.
⬇ 11.19 → Significant difference
冥
冥 冥 冥
共x 3 x4 兲2 SSw 1 1 兺共ni 1兲 n 3 n4
⬇ 2.91 → No difference
⬇ 12.70 → Significant difference
⬇ 2.17 → No difference
⬇ 25.45 → Significant difference
冥
Mean
Size
Pop 1
106.26
9
Pop 2
116.71
11
Pop 3
82.49
9
Pop 4
70.07
10
⬇ 45.36 → Significant difference
SSw ⬇ 57,600.361 兺共ni 1兲 F0 2.255 → CVScheffé 2.255共4 1兲 6.765
共x 1 x2兲2 ⬇ 0.329 → No difference SSw 1 1 兺共ni 1兲 n1 n2
冤
冥
共x 1 x 3兲2 ⬇ 1.545 → No difference SSw 1 1 兺共ni 1兲 n1 n3
冤
冥
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共x 1 x 4兲2 ⬇ 3.769 → No difference SSw 1 1 兺共ni 1兲 n1 n4
冤
冥
共x 2 x 3兲2 ⬇ 3.522 → No difference SSw 1 1 兺共ni 1兲 n2 n3
冤
冥
共x2 x4兲2 ⬇ 6.923 → Significant difference SSw 1 1 兺共ni 1兲 n2 n4
冤
冥
共x 3 x 4兲2 ⬇ 0.444 → No difference SSw 1 1 兺共ni 1兲 n3 n4
冤
冥
CHAPTER 10 REVIEW EXERCISE SOLUTIONS 1. Claimed distribution: Category
Distribution
0 1–3 4–9 10+
16% 46% 25% 13%
H0: Distribution of health care visits is as shown in table above. Ha: Distribution of health care visits differs from the claimed distribution.
20 7.815 Category 0 1–3 4–9 10+
Distribution
Observed
Expected
冇O ⴚ E冈2 E
16% 46% 25% 13%
99 376 167 92
117.44 337.64 183.50 95.42
2.895 4.368 1.484 0.123
734
8.860
20 8.860 Reject H0. There is enough evidence at the 5% significance level to reject the claimed or expected distribution. 2. Claimed distribution: Advertisers
Distribution
Shortgame Approach Driver Putting
65% 22% 9% 4%
H0: Distribution of advertisers is as shown in table above. Ha: Distribution of advertisers differs from the claimed distribution.
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20 6.251 Advertisers
Distribution
Observed
Expected
Shortgame Approach Driver Putting
65% 22% 9% 4%
276 99 42 18
282.75 95.70 39.15 17.40
435
(O ⴚ E )2 E 0.161 0.114 0.207 0.021 0.503
2 0.503 Fail to reject H0. There is not enough evidence at the 10% significance level to reject the claimed or expected distribution. 3. Claimed distribution: Days
Distribution
0 1–5 6–10 11–15 16–20 21+
19% 30% 30% 12% 3% 6%
H0: Distribution of days is as shown in the table above. Ha: Distribution of days differs from the claimed distribution.
20 9.236 Days
Distribution
Observed
Expected
(O ⴚ E )2 E
0 1–5 6–10 11–15 16–20 21+
19% 30% 30% 12% 3% 6%
165 237 245 88 19 46
152 240 240 96 24 48
1.112 0.038 0.104 0.667 1.042 0.083
800
3.045
2 ⬇ 3.045 Fail to reject H0. There is not enough evidence at the 10% significance level to support the claim. 4. Claimed distribution: Age 21–29 30–39 40–49 50–59 60+
Distribution 20.5% 21.7% 18.1% 17.3% 22.4%
H0: Distribution of ages is as shown in table above. Ha: Distribution of ages differs from the claimed distribution.
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CHAPTER 10

CHISQUARE TESTS AND THE FDISTRIBUTION
20 13.277 Age
Distribution
Observed
Expected
(O ⴚ E )2 E
20.5% 21.7% 18.1% 17.3% 22.4%
45 128 244 224 359
205 217 181 173 224
124.878 36.502 21.928 15.035 81.362
21–29 30–39 40–49 50–59 60+
1000
279.705
2 ⬇ 279.705 Reject H0. There is enough evidence at the 1% significance level to support the claim that the distribution of ages of the jury differs from the age distribution of available jurors. 5. (a) Expected frequencies: HS– did not complete
HS complete
College 1–3 years
College 4+ years
Total
25–44 45+
663.49 856.51
1440.47 1859.53
1137.54 1468.46
1237.5 1597.5
4479 5782
Total
1520
3300
2606
2835
10,261
(b) H0 : Education is independent of age. Ha: Education is dependent on age. d.f. 3 20 6.251
2
兺
共O E兲 2 ⬇ 66.128 E
Reject H0. (c) There is enough evidence at the 10% significance level to conclude that education level of people in the United States and their age are dependent. 6. (a) Expected frequencies: Types of vehicles owned Gender
Car
Truck
SUV
Van
Total
Male Female
94.25 100.75
82.65 88.35
50.75 54.25
4.35 4.65
232 248
195
171
105
9
480
Total
(b) H0: Type of vehicle is independent of gender. Ha: Type of vehicle is dependent on gender.
20 7.815 2
兺
共O E兲2 8.403 E
Reject H0. (c) There is enough evidence at the 5% significance level to conclude that type of vehicle owned is dependent on gender. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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7. (a) Expected frequencies: Age Group Gender
16 –20
21–30
31– 40
41–50
51– 60
61+
Total
Male Female
143.76 71.24
284.17 140.83
274.14 135.86
237.37 117.63
147.10 72.90
43.46 21.54
1130 560
215
425
410
355
220
65
1690
Total
(b) H0 : Gender is independent of age. Ha: Gender and age are dependent. d.f. 5 20 11.071
2
兺
共O E兲2 9.951 E
Fail to reject H0. (c) There is not enough evidence at the 10% significance level to conclude that gender and age group are dependent. 8. (a) Expected frequencies: Time of day Gender
12 A.M.– 5:59 A.M.
6 A.M.– 11:59 A.M.
12 P.M.– 5:59 P.M.
6 P.M.– 11:59 P.M.
Total
Male Female
582.68 326.32
612.81 343.19
967.93 542.07
918.57 514.43
3082 1726
909
956
1510
1432
4808
Total
(b) H0: Type of vehicle is independent of gender. Ha: Type of vehicle is dependent on gender.
20 6.251 2 36.742 Reject H0. (c) There is enough evidence at the 10% significance level to conclude that time and gender are dependent. 9. F0 ⬇ 2.295
10. F0 4.71
11. F0 2.39
12. F0 2.01
13. H0: 21 22 (claim); Ha: 21 > 22 d.f.N 15 d.f.D 20 F0 3.09; Reject H0 if F > 3.09. F
s 21 653 ⬇ 2.419 s 22 270
Fail to reject H0. There is not enough evidence to reject the claim.
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CHAPTER 10

CHISQUARE TESTS AND THE FDISTRIBUTION
14. H0: 21 22 ; Ha: 21 22 (claim) d.f.N 5 d.f.D 10 F0 3.33; Reject H0 if F > 3.33. F
s 21 112,676 ⬇ 2.273 s 22 49,572
Fail to reject H0. There is not enough evidence at the 10% significance level to support the claim. 15. Population 1: Garfield County Population 2: Kay County H0: 21 22 ; Ha : 21 > 22 (claim) d.f.N 20 d.f.D 15 F0 1.92; Reject H0 if F > 1.92. F
s 21 共0.76兲2 ⬇ 1.717 s 22 共0.58兲2
Fail to reject H0. There is not enough evidence at the 10% significance level to support the claim that the variation in wheat production is greater in Garfield County than in Kay County. 16. Population 1: Nontempered Population 2: Tempered H0: 21 22 ; Ha: 21 > 22 (claim) d.f.N 8 d.f.D 8 F0 3.44; Reject H0 if F > 3.44. F
s 21 共25.4兲2 ⬇ 3.759 s 22 共13.1兲2
Reject H0. There is enough evidence at the 5% significance level to support the claim that the yield strength of the nontempered couplings is more variable than that of the tempered couplings. 17. Population 1: Male → s 21 18,486.26 Population 2: Female → s 22 12,102.78 H0: 21 22 ; Ha: 21 22 (claim) d.f.N 13 d.f.D 8 F0 6.94; Reject H0 if F > 6.94. F
s 21 18,486.26 ⬇ 1.527 s 22 12,102.78
Fail to reject H0. There is not enough evidence at the 1% significance level to support the claim that the test score variance for females is different than that for males. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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18. Population 1: Current → s 21 0.00146 Population 2: New → s 22 0.00050 H0: 21 22 ; Ha: 21 > 22 (claim) d.f.N 11 d.f.D 11 F0 2.82; Reject H0 if F > 2.82. F
s 21 0.00146 ⬇ 2.92 s 22 0.00050
Reject H0. There is enough evidence at the 5% significance level to support the claim that the new mold produces inserts that are less variable in diameter than the current mold. 19. H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) d.f.N k 1 3 d.f.D N k 28 F0 2.29; Reject H0 if F > 2.29. Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
512.457 562.162
3 28
170.819 20.077
F 8.508
F ⬇ 8.508 Reject H0. There is enough evidence at the 10% significance level to conclude that the mean residential energy expenditures are not the same for all four regions. 20. H0: 1 2 3 4 Ha: At least one mean is different from the others. (claim) d.f.N k 1 3 d.f.D N k 20 F0 3.10; Reject H0 if F > 3.10. Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
461,446,125 2,829,846,458
3 20
153,815,375 141,492,323
F 1.087
F ⬇ 1.087 Fail to reject H0. There is not enough evidence a the 5% significance level to conclude that the average annual incomes are not the same for the four regions.
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CHAPTER 10

CHISQUARE TESTS AND THE FDISTRIBUTION
CHAPTER 10 QUIZ SOLUTIONS 1. (a) Population 1: San Jose → s 21 ⬇ 430.084 Population 2: Dallas → s 22 ⬇ 120.409 H0: 21 22 ; Ha: 21 22 (claim) (b) 0.01 (cd) d.f.N 12 d.f.D 15 F0 4.25; Reject H0 if F > 4.25. (e) F
s 21 430.084 ⬇ 3.57 s 22 120.409
(f) Fail to reject H0. (g) There is not enough evidence at the 1% significance level to conclude that the variances in annual wages for San Jose, CA and Dallas, TX are different. 2. (a) H0: 1 2 3 (claim) Ha: At least one mean is different from the others. (b) 0.10 (cd) d.f.N k 1 2 d.f.D N k 40 F0 2.44; Reject H0 if F > 2.44. Variation
Sum of Squares
Degrees of Freedom
Between Within
5021.896 8224.121
2 40
Mean Squares
F
2510.948 12.213 205.603
(e) F ⬇ 12.21 (f) Reject H0. (g) There is enough evidence at the 10% significance level to reject the claim that the mean annual wages for the three cities are not all equal. 3. (a) Claimed distribution: Education Not a HS graduate HS graduate Some college, no degree Associate’s degree Bachelor’s degree Advanced degree
25 & Over 14.8% 32.2% 16.8% 8.6% 18.1% 9.5%
H0 : Distribution of educational achievement for people in the United States ages 35–44 is as shown in table above. Ha: Distribution of educational achievement for people in the United States ages 35–44 differs from the claimed distribution. (claim) (b) 0.01 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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(cd) 20 15.086; Reject H0 if 2 > 15.086. (e) Education
25 & Over
Observed
Expected
(O ⴚ E )2 E
14.8% 32.2% 16.8% 8.6% 18.1% 9.5%
35 95 51 30 61 29
44.548 96.922 50.568 25.886 54.481 28.595
2.046 0.038 0.004 0.654 0.780 0.006
Not a HS graduate HS graduate Some college, no degree Associate’s degree Bachelor’s degree Advanced degree
301
3.528
2 3.528 (f) Fail to reject H0. (g) There is not enough evidence at the 1% significance level to conclude that the distribution of educational achievement for people in the United States ages 35–44 differs from the claimed or expected distribution. 4. (a) Claimed distribution: Education
25 & Over
Not a HS graduate HS graduate Some college, no degree Associate’s degree Bachelor’s degree Advanced degree
14.8% 32.2% 16.8% 8.6% 18.1% 9.5%
H0 : Distribution of educational achievement for people in the United States ages 65–74 is as shown in table above. Ha: Distribution of educational achievement for people in the United States ages 65–74 differs from the claimed distribution. (claim) (b) 0.05 (c) 20 11.071 (d) Reject H0 if 2 > 11.071. (e) Education Not a HS graduate HS graduate Some college, no degree Associate’s degree Bachelor’s degree Advanced degree
25 & Over
Observed
Expected
(O ⴚ E )2 E
14.8% 32.2% 16.8% 8.6% 18.1% 9.5%
91 151 58 23 50 33
60.088 130.730 68.208 34.916 73.486 38.570
15.903 3.142 1.528 4.067 7.506 0.804
406
32.950
2 32.950 (f ) Reject H0. (g) There is enough evidence at the 5% significance level to conclude that the distribution of educational achievement for people in the United States ages 65–74 differs from the claimed or expected distribution.
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CHAPTER
Nonparametric Tests
11 11.1 THE SIGN TEST
11.1 Try It Yourself Solutions 1a. H0 : median ⱕ 2500; Ha: median > 2500 (claim) b. ␣ ⫽ 0.025 c. n ⫽ 22 d. The critical value is 5. e. x ⫽ 10 f. Fail to reject H0 . g. There is not enough evidence to support the claim. 2a. H0 : median ⫽ $134,500 (claim); Ha: median ⫽ $134,500 b. ␣ ⫽ 0.10 c. n ⫽ 81 d. The critical value is z0 ⫽ ⫺1.645. e. x ⫽ 30 共x ⫹ 0.5兲 ⫺ 0.5共n兲 共30 ⫹ 0.5兲 ⫺ 0.5共81兲 ⫺10 z⫽ ⫽ ⫽ ⫽ ⫺2.22 4.5 冪n 冪81 2 2 f. Reject H0 . g. There is enough evidence to reject the claim. 3a. H0 : The number of colds will not decrease. Ha: The number of colds will decrease. (claim) b. ␣ ⫽ 0.05 c. n ⫽ 11 d. The critical value is 2. e. x ⫽ 2 f. Reject H0 . g. There is enough evidence to support the claim.
11.1 EXERCISE SOLUTIONS 1. A nonparametric test is a hypothesis test that does not require any specific conditions concerning the shape of populations or the value of any population parameters. A nonparametric test is usually easier to perform than its corresponding parametric test, but the nonparametric test is usually less efficient. 325 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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2. Identify the claim and state H0 and Ha. Identify the level of significance and sample size. Find the critical value using Table 8 (if n ⱕ 25) or Table 4 (n > 25). Calculate the test statistic. Make a decision and interpret in the context of the problem. 3. (a) H0: median ⱕ $300; Ha: median > $300 (claim) (b) Critical value is 1. (c) x ⫽ 5 (d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that the median amount of new credit card charges for the previous month was more than $300. 4. (a) H0: median ⫽ 83 (claim); Ha: median ⫽ 83 (b) The critical value is 1. (c) x ⫽ 4 (d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to reject the claim that the daily median temperature for the month of July in Pittsburgh is 83⬚ Fahrenheit. 5. (a) H0: median ⱕ $210,000 (claim); Ha: median > $210,000 (b) Critical value is 1. (c) x ⫽ 3 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to reject the claim that the median sales price of new privately owned onefamily homes sold in the past year is $210,000 or less. 6. (a) H0: median ⫽ 66 (claim); Ha: median ⫽ 66 (b) The critical value is 3. (c) x ⫽ 0 (d) Reject H0. (e) There is enough evidence at the 1% significance level to reject the claim that the daily median temperature for the month of January in San Diego is 66⬚F. 7. (a) H0: median ⱖ $2200 (claim); Ha: median < $2200 (b) Critical value is z0 ⫽ ⫺2.05. (c) x ⫽ 44 z⫽
共x ⫹ 0.5兲 ⫺ 0.5共n兲 共44 ⫹ 0.5兲 ⫺ 0.5共104兲 ⫺7.5 ⫽ ⫽ ⬇ ⫺1.47 5.099 冪n 冪104 2 2
(d) Fail to reject H0. (e) There is not enough evidence at the 2% significance level to reject the claim that the median amount of credit card debt for families holding such debts is at least $2200.
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CHAPTER 11

NONPARAMETRIC TESTS
8. (a) H0: median ⱖ $50,000; Ha : median < $50,000 (claim) (b) The critical value is z0 ⫽ ⫺1.96. (c) x ⫽ 24 z⫽
共x ⫹ 0.5兲 ⫺ 0.5共n兲 共24 ⫹ 0.5兲 ⫺ 0.5共70兲 ⫺10.5 ⫽ ⫽ ⬇ ⫺2.51 4.183 冪n 冪70 2 2
(d) Reject H0. (e) There is enough evidence at the 2.5% significance level to support the claim that the median amount of financial debt for families holding such debts is less than $50,000. 9. (a) H0: median ⱕ 30; Ha: median > 30 (claim) (b) Critical value is 2. (c) x ⫽ 4 (d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that the median age of recipients of engineering doctorates is greater than 30 years. 10. (a) H0: median ⱖ 32; Ha: median < 32 (claim) (b) The critical value is 5. (c) x ⫽ 5 (d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that the median age of recipients of biological science doctorates is less than 32 years. 11. (a) H0: median ⫽ 4 (claim); Ha: median ⫽ 4 (b) Critical value is z0 ⫽ ⫺1.96. (c) x ⫽ 13 z⫽
共x ⫹ 0.5兲 ⫺ 0.5共n兲 共13 ⫹ 0.5兲 ⫺ 0.5共33兲 ⫺3 ⫽ ⫽ ⬇ ⫺1.04 2.872 冪n 冪33 2 2
(d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to reject the claim that the median number of rooms in renteroccupied units is 4. 12. (a) H0: median ⫽ 1000 (claim); Ha: median ⫽ 1000 (b) The critical value is 5. (c) x ⫽ 5 (d) Reject H0. (e) There is enough evidence at the 10% significance level to reject the claim that the median square footage of renteroccupied units is 1000 square feet.
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13. (a) H0: median ⫽ $12.16 (claim); Ha: median ⫽ $12.16 (b) Critical value is z0 ⫽ ⫺2.575. (c) x ⫽ 16 z⫽
共x ⫹ 0.5兲 ⫺ 0.5共n兲 共16 ⫹ 0.5兲 ⫺ 0.5共39兲 ⫺3 ⫽ ⫽ ⬇ ⫺0.961 3.1225 冪n 冪39 2 2
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to reject the claim that the median hourly earnings of male workers paid hourly rates is $12.16. 14. (a) H0: median ⱕ $10.31 (claim); Ha: median > $10.31 (b) The critical value is 5. (c) x ⫽ 9 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to reject the claim that the median hourly earnings of female workers paid hourly rates is at most $10.31. 15. (a) H0: The lower back pain intensity scores have not decreased. Ha: The lower back pain intensity scores have decreased. (claim) (b) Critical value is 1. (c) x ⫽ 0 (d) Reject H0. (e) There is enough evidence at the 5% significance level to conclude that the lower back pain intensity scores were reduced after accupunture. 16. (a) H0: The lower back pain intensity scores have not decreased. Ha: The lower back pain intensity scores have decreased. (claim) (b) The critical value is 2. (c) x ⫽ 4 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to support the claim that the lower back pain intensity scores have decreased. 17. (a) H0: The SAT scores have not improved. Ha: The SAT scores have improved. (claim) (b) Critical value is 2. (c) x ⫽ 4 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to support the claim that the verbal SAT scores improved.
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CHAPTER 11

NONPARAMETRIC TESTS
18. (a) H0: The SAT scores have not improved. Ha: The SAT scores have improved. (claim) (b) The critical value is 1. (c) x ⫽ 3 (d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that the verbal SAT scores have improved. 19. (a) H0: The proportion of adults who prefer unplanned travel activities is equal to the proportion of adults who prefer planned travel activities. (claim) Ha: The proportion of adults who prefer unplanned travel activities is not equal to the proportion of adults who prefer planned travel activities. Critical value is 3. ␣ ⫽ 0.05 x⫽5 Fail to reject H0. (b) There is not enough evidence at the 5% significance level to reject the claim that the proportion of adults who prefer unplanned travel activities is equal to the proportions of adults who prefer planned travel activities. 20. (a) H0: The proportion of adults who contact their parents by phone weekly is equal to the proportion of adults who contact their parents by phone daily. (claim) Ha: The proportion of adults who contact their parents by phone weekly is not equal to the proportion of adults who contact their parents by phone daily. The critical value is 6. 共␣ ⫽ 0.05兲. x⫽9 Fail to reject H0. (b) There is not enough evidence at the 5% significance level to reject the claim that the proportion of adults who contact their parents by phone weekly is equal to the proportion of adults who contact their parents by phone daily. 21. (a) H0: median ⱕ $585 (claim); Ha: median > $585 (b) Critical value is z0 ⫽ 2.33. (c) x ⫽ 29 z⫽
共x ⫺ 0.5兲 ⫺ 0.5共n兲 共29 ⫺ 0.5兲 ⫺ 0.5共47兲 5 ⫽ ⫽ ⬇ 1.46 3.428 冪n 冪47 2 2
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to reject the claim that the median weekly earnings of female workers is less than or equal to $585.
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22. (a) H0: median ⱕ $720; Ha: median > $720 (claim) (b) The critical value is z0 ⫽ 2.33. (c) x ⫽ 45 z⫽
共x ⫺ 0.5兲 ⫺ 0.5共n兲 共45 ⫺ 0.5兲 ⫺ 0.5共68兲 10.5 ⫽ ⫽ ⬇ 2.55 4.123 冪n 冪68 2 2
(d) Reject H0. (e) There is enough evidence at the 1% significance level to support the claim that the median weekly earnings of male workers is greater than $720. 23. (a) H0: median ⱕ 25.5; Ha: median > 25.5 (claim) (b) Critical value is z0 ⫽ 1.645. (c) x ⫽ 38 共x ⫹ 0.5兲 ⫺ 0.5共n兲 共38 ⫹ 0.5兲 ⫺ 0.5共60兲 8.5 z⫽ ⫽ ⫽ ⬇ 1.936 3.873 冪n 冪60 2 2 (d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that the median age of firsttime brides is greater than 25.5 years. 24. (a) H0: median ⱕ 27 (claim); Ha: median > 27 (b) The critical value is z0 ⫽ 1.645. (c) x ⫽ 33 z⫽
共x ⫺ 0.5兲 ⫺ 0.5共n兲 共33 ⫺ 0.5兲 ⫺ 0.5共56兲 4.5 ⫽ ⫽ ⬇ 1.203 3.742 冪n 冪56 2 2
(d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to reject the claim that the median age of firsttime grooms is less than or equal to 27.
11.2 THE WILCOXON TESTS
11.2 Try It Yourself Solutions 1a. H0 : The water repellent is not effective. Ha: The water repellent is effective. (claim) b. ␣ ⫽ 0.01 c. n ⫽ 11 d. Critical value is 5.
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CHAPTER 11
e.
No repellent
Repellent applied
Difference
Absolute value
Rank
Signed rank
8 7 7 4 6 10 9 5 9 11 8 4
15 12 11 6 6 8 8 6 12 8 14 8
⫺7 ⫺5 ⫺4 ⫺2 0 2 1 ⫺1 ⫺3 3 ⫺6 ⫺4
7 5 4 2 0 2 1 1 3 3 6 4
11 9 7.5 3.5 — 3.5 1.5 1.5 5.5 5.5 10 7.5
⫺ 11 ⫺9 ⫺ 7.5 ⫺ 3.5 — 3.5 1.5 ⫺ 1.5 ⫺ 5.5 5.5 ⫺ 10 ⫺ 7.5

NONPARAMETRIC TESTS
Sum of negative ranks ⫽ ⫺55.5 Sum of positive ranks ⫽ 10.5 ws ⫽ 10.5 f. Fail to reject H0 . g. There is not enough evidence at the 1% significance level to support the claim. 2a. H0: There is no difference in the claims paid by the companies. Ha: There is a difference in the claims paid by the companies. (claim) b. ␣ ⫽ 0.05 c. The critical values are z0 ⫽ ± 1.96. d. n1 ⫽ 12 and n2 ⫽ 12 e.
Ordered data
Sample
Rank
Ordered data
Sample
Rank
1.7 1.8 2.2 2.5 3.0 3.0 3.4 3.9 4.1 4.4 4.5 4.7
B B B A A B B A B B A B
1 2 3 4 5.5 5.5 7 8 9 10 11 12
5.3 5.6 5.8 6.0 6.2 6.3 6.5 7.3 7.4 9.9 10.6 10.8
B B A A A A A B A A A B
13 14 15 16 17 18 19 20 21 22 23 24
R ⫽ sum ranks of company B ⫽ 120.5 f. R ⫽
R ⫽ z⫽
n1共n1 ⫹ n2 ⫹ 1兲 12共12 ⫹ 12 ⫹ 1兲 ⫽ ⫽ 150 2 2
冪n n 共n 12⫹ n 1 2
1
2
⫹ 1兲
⫽
冪共12兲共12兲共1212⫹ 12 ⫹ 1兲 ⬇ 17.321
R ⫺ R 120.5 ⫺ 150 ⫽ ⬇ ⫺1.703 R 17.321
g. Fail to reject H0 . h. There is not enough evidence to conclude that there is a difference in the claims paid by both companies. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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11.2 EXERCISE SOLUTIONS 1. The Wilcoxon signedrank test is used to determine whether two dependent samples were selected from populations having the same distribution. The Wilcoxon rank sum test is used to determine whether two independent samples were selected from populations having the same distribution. 2. The sample size of both samples must be at least 10. 3. (a) H0: There is no reduction in diastolic blood pressure. (claim) Ha: There is a reduction in diastolic blood pressure. (b) Wilcoxon signedrank test (c) Critical value is 10. (d) ws ⫽ 17 (e) Fail to reject H0. (f ) There is not enough evidence at the 1% significance level to reject the claim that there was no reduction in diastolic blood pressure. 4. (a) H0: There is no difference in salaries. (claim) Ha: There is a difference in salaries. (b) Wilcoxon rank sum test (c) The critical value is z0 ⫽ ± 1.645. (d) R ⫽ 127 n 共n ⫹ n2 ⫹ 1兲 10共10 ⫹ 10 ⫹ 1兲 R ⫽ 1 1 ⫽ ⫽ 105 2 2
R ⫽ z⫽
冪n n 共n 12⫹ n 1 2
1
2
⫹ 1兲
⫽
冪共10兲共10兲共1012⫹ 10 ⫹ 1兲 ⫽ 13.229
R ⫺ R 127 ⫺ 105 ⫽ ⬇ 1.663 R 13.229
(e) Reject H0. (f) There is enough evidence at the 10% significance level to reject the claim that there is no difference in the salaries. 5. (a) H0: There is no difference in the earnings. Ha: There is a difference in the earnings. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0 ⫽ ± 1.96.
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(d) R ⫽ 58 n 共n ⫹ n2 ⫹ 1兲 11共11 ⫹ 10 ⫹ 1兲 ⫽ ⫽ 110 R ⫽ 1 1 2 2
R ⫽ z⫽
冪n n 共n 12⫹ n 1 2
1
2
⫹ 1兲
⫽
冪共11兲共10兲共1112⫹ 10 ⫹ 1兲 ⬇ 14.201
R ⫺ R 58 ⫺ 110 ⫽ ⬇ ⫺3.66 R 14.201
(e) Reject H0. (f ) There is enough evidence at the 5% significance level to support the claim that there is a difference in the earnings. 6. (a) H0: There is no difference in the number of months mothers breastfeed their babies. (claim) Ha: There is a difference in the number of months mothers breastfeed their babies. (b) Wilcoxon rank sum test (c) The critical value is z0 ⫽ ± 2.575. (d) R ⫽ 104 n 共n ⫹ n2 ⫹ 1兲 11共11 ⫹ 12 ⫹ 1兲 R ⫽ 1 1 ⫽ ⫽ 132 2 2
R ⫽ z⫽
冪n n 共n 12⫹ n 1 2
1
2
⫹ 1兲
⫽
冪共11兲共12兲共1112⫹ 12 ⫹ 1兲 ⫽ 16.248
R ⫺ R 104 ⫺ 132 ⫽ ⬇ ⫺1.723 R 16.248
(e) Fail to reject H0. (f) There is not enough evidence at the 1% significance level to reject the claim that there is no difference in the number of months mothers breastfeed their babies. 7. (a) H0: There is not a difference in salaries. Ha: There is a difference in salaries. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0 ⫽ ± 1.96. (d) R ⫽ 117
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 12共12 ⫹ 12 ⫹ 1兲 ⫽ ⫽ 150 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪共12兲共12兲共1212⫹ 12 ⫹ 1兲 ⬇ 17.321
R ⫺ R 117 ⫺ 150 ⫽ ⬇ ⫺1.91 R 17.321
(e) Fail to reject H0. (f ) There is not enough evidence at the 5% significance level to support the claim that there is a difference in salaries.
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8. (a) H0: The new drug does not affect the number of headache hours. Ha: The new drug does affect the number of headache hours. (claim) (b) Wilcoxon signedrank test (c) The critical value is 2. (d) Before
After
Difference
0.8 2.4 2.8 2.6 2.7 0.9 1.2
1.6 1.3 1.6 1.4 1.5 1.6 1.7
⫺0.8 1.1 1.2 1.2 1.2 ⫺0.7 ⫺0.5
Absolute value 0.8 1.1 1.2 1.2 1.2 0.7 0.5
Rank
Signed rank
3 4 6 6 6 2 1
⫺3 4 6 6 6 ⫺2 ⫺1
The sum of the negative ranks is ⫺3 ⫹ 共⫺2兲 ⫹ 共⫺1兲 ⫽ ⫺6. The sum of the positive ranks is 4 ⫹ 6 ⫹ 6 ⫹ 6 ⫽ 22. Because ⫺6 < 22 , the test statistic is ws ⫽ 6.
ⱍ ⱍ ⱍ ⱍ
(e) Fail to reject H0. (f) There is not enough evidence at the 5% significance level to conclude that the new drug affects the number of headache hours. 9. H0: The fuel additive does not improve gas mileage. Ha: The fuel additive does improve gas mileage. (claim) Critical value is z0 ⫽ 1.282. ws ⫽ 43.5
z⫽
ws ⫺
n共n ⫹ 1兲 4
⫽
43.5 ⫺
32共32 ⫹ 1兲 4
冪n共n ⫹ 124兲共2n ⫹ 1兲 冪32共32 ⫹ 124兲关共2兲32 ⫹ 1兴
⫽
⫺220.5 ⬇ ⫺4.123 冪2860
Note: n ⫽ 32 because one of the differences is zero and should be discarded. Reject H0. There is enough evidence at the 10% level to conclude that the gas mileage is improved. 10. H0: The fuel additive does not improve gas mileage. Ha: The fuel additive does improve gas mileage. (claim) The critical value is z0 ⫽ ⫺1.645. ws ⫽ 0
z⫽
ws ⫺
n共n ⫹ 1兲 4
冪n共n ⫹ 124兲共2n ⫹ 1兲
⫽
0 ⫺ 264 ⫽ ⫺4.937 冪2860
Reject H0. There is enough evidence at the 5% level to conclude that the fuel additive improves gas mileage.
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11.3 THE KRUSKALWALLIS TEST
11.3 Try It Yourself Solutions 1a. H0 : There is no difference in the salaries in the three states. Ha: There is a difference in the salaries in the three states. (claim) b. ␣ ⫽ 0.10 c. d.f. ⫽ k ⫺ 1 ⫽ 2 d. Critical value is 20 ⫽ 4.605; Reject H0 if 2 > 4.605. e.
Ordered data
State
86.85 87.25 87.70 89.65 89.75 89.92 91.17 91.55 92.85 93.12 93.76 93.92 94.42 94.45 94.55
R1 ⫽ 127 f. H ⫽ ⫽
MI NC CO MI CO NC CO CO CO CO MI MI MI MI CO
R2 ⫽ 148
12 N共N ⫹ 1兲
冢
Rank
Ordered data
State
Rank
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
94.72 94.75 95.10 95.36 96.02 96.24 96.31 97.35 98.21 98.34 98.99 100.27 105.77 106.78 110.99
NC NC MI NC NC CO MI CO MI NC CO NC NC MI NC
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
R3 ⫽ 190
冣
R 21 R2 R2 ⫹ 2 ⫹ 3 ⫺ 3共N ⫹ 1兲 n1 n2 n3
12 共127兲2 共148兲2 共190兲2 ⫹ ⫹ ⫺ 3共30 ⫹ 1兲 ⫽ 2.655 30共30 ⫹ 1兲 10 10 10
冢
冣
g. Fail to reject H0. h. There is not enough evidence to support the claim.
11.3 EXERCISE SOLUTIONS 1. Each sample must be randomly selected and the size of each sample must be at least 5. 2. The KruskalWallis test is always a righttailed test because the null hypothesis is only rejected when H is significantly large. 3. (a) H0: There is no difference in the premiums. Ha: There is a difference in the premiums. (claim) (b) Critical value is 5.991. (c) H ⬇ 13.091 (d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that the distributions of the annual premiums in Arizona, Florida, and Louisiana are different. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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4. (a) H0: There is no difference in the premiums. Ha: There is a difference in the premiums. (claim) (b) The critical value is 5.991. (c) H ⬇ 9.841 (d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that the distributions of the annual premiums in the three states are different. 5. (a) H0: There is no difference in the salaries. Ha: There is a difference in the salaries. (claim) (b) Critical value is 6.251. (c) H ⬇ 1.024 (d) Fail to reject H0. (e) There is not enough evidence at the 10% significance level to support the claim that the distributions of the annual salaries in the four states are different. 6. (a) H0: There is no difference in the salaries. Ha: There is a difference in the salaries. (claim) (b) The critical value is 6.251. (c) H ⬇ 15.713 (d) Reject H0. (e) There is enough evidence at the 10% significance level to support the claim that the distributions of the annual salaries in the four states are different. 7. (a) H0: There is no difference in the number of days spent in the hospital. Ha: There is a difference in the number of days spent in the hospital. (claim) The critical value is 11.345. H ⫽ 1.51; Fail to reject H0 . (b) Variation
Sum of squares
Degrees of freedom
Mean Squares
Between Within
9.17 194.72
3 33
3.06 5.90
F 0.52
For ␣ ⫽ 0.01, the critical value is about 4.45. Because F ⫽ 0.52 is less than the critical value, the decision is to fail to reject H0. There is not enough evidence to support the claim. (c) Both tests come to the same decision, which is that there is not enough evidence to support the claim that there is a difference in the number of days spent in the hospital.
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CHAPTER 11

NONPARAMETRIC TESTS
8. (a) H0: There is no difference in the mean energy consumptions. Ha: There is a difference in the mean energy consumptions. (claim) The critical value is 11.345. H ⬇ 16.26 Reject H0. There is enough evidence to support the claim. (b) Variation
Sum of Squares
Degrees of Freedom
Mean Squares
Between Within
32,116 45,794
3 35
10,705 1308
F 8.18
For ␣ ⫽ 0.01, the critical value is about 4.41. Because F ⫽ 8.18 is greater than the critical value, the decision is to reject H0. There is enough evidence to support the claim. (c) Both tests come to the same decision, which is that the mean energy consumptions are different.
11.4 RANK CORRELATION
11.4 Try It Yourself Solutions 1a. H0 : s ⫽ 0; Ha : s ⫽ 0 b. ␣ ⫽ 0.05 c. Critical value is 0.700. d.
Oat
Rank
Wheat
Rank
d
d2
1.5 3 1.5 6 8 4.5 4.5 7 9
2.65 2.48 2.62 2.78 3.56 3.40 3.40 4.25 4.25
3 1 2 4 8 5.5 5.5 7 9
⫺1.5 2 ⫺0.5 2 0 ⫺1 ⫺1 0 0
2.25 4 0.25 4 0 1 1 0 0
1.10 1.12 1.10 1.59 1.81 1.48 1.48 1.63 1.85
兺 ⫽ 12.5
兺d 2 ⫽ 12.5 e. rs ⫽ 1 ⫺
6兺d 2 ⫽ 0.896 n共n 2 ⫺ 1兲
f. Reject H0 . g. There is enough evidence to conclude that a significant correction exists.
11.4 EXERCISE SOLUTIONS 1. The Spearman rank correlation coefficient can (1) be used to describe the relationship between linear and nonlinear data, (2) be used for data at the ordinal level, and (3) is easier to calculate by hand than the Pearson coefficient.
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2. The ranks of corresponding data pairs are identical. The ranks of corresponding data pairs are in reverse order. The ranks of corresponding data pairs have no relationship. 3. (a) H0: s ⫽ 0; Ha: s ⫽ 0 (claim) (b) Critical value is 0.929. (c) 兺d 2 ⫽ 8 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.857 n共n2 ⫺ 1兲
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that there is a correlation between debt and income in the farming business. 4. (a) H0: s ⫽ 0; Ha: s ⫽ 0 (claim) (b) The critical value is 0.618. (c) 兺d 2 ⫽ 65.6 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.702 n共n2 ⫺ 1兲
(d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that there is a correlation between the overall score and price. 5. (a) H0: s ⫽ 0; Ha: s ⫽ 0 (claim) (b) The critical value is 0.881. (c) 兺d 2 ⫽ 66.5 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.208 n共n2 ⫺ 1兲
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that there is a correlation between the overall score and price. 6. (a) H0: s ⫽ 0; Ha: s ⫽ 0 (claim) (b) Critical value is 0.497. (c) 兺d 2 ⫽ 165.5 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.421 n共n2 ⫺ 1兲
(d) Fail to reject H0. (e) There is not enough evidence at the 10% significance level to support the claim that there is a correlation between overall score and price.
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CHAPTER 11

NONPARAMETRIC TESTS
7. H0: s ⫽ 0; Ha: s ⫽ 0 (claim) Critical value is 0.700. 兺d 2 ⫽ 121.5 rs ⫽ 1 ⫺
6兺d 2 ⬇ ⫺0.013 n共n2 ⫺ 1兲
Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that there is a correlation between science achievement scores and GNP. 8. H0: s ⫽ 0; Ha: s ⫽ 0 (claim) The critical value is 0.700. 兺d 2 ⫽ 120 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.000 n共n2 ⫺ 1兲
Fail to reject H0. There is not enough evidence at the 5% significance level to conclude that there is a correlation between mathematics achievement scores and GNP. 9. H0: s ⫽ 0; Ha: s ⫽ 0 (claim) Critical value is 0.700. 兺d 2 ⫽ 9.5 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.921 n共n2 ⫺ 1兲
Reject H0. There is enough evidence at the 5% significance level to conclude that there is a correlation between science and mathematics achievement scores. 10. There is not enough evidence at the 5% significance level to conclude that there is a significant correlation between eighth grade science achievement scores and the GNP of a country, or between eighth grade math achievement scores and the GNP of a country. However, there is enough evidence at the 5% significance level to conclude there is a significant correlation. 11. H0: s ⫽ 0; Ha: s ⫽ 0 (claim) The critical value is ⫽
±z 冪n ⫺ 1
⫽
± 1.96 冪33 ⫺ 1
⬇ ± 0.346.
兺d 2 ⫽ 6673 rs ⫽ 1 ⫺
6兺d 2 ⬇ ⫺0.115 n共n2 ⫺ 1兲
Fail to reject H0. There is not enough evidence to support the claim. 12. H0: s ⫽ 0; Ha: s ⫽ 0 (claim) Critical value ⫽
±z 冪n ⫺ 1
⫽
± 1.96 冪34 ⫺ 1
⬇ ± 0.341.
兺d 2 ⫽ 7310.5 rs ⫽ 1 ⫺
6兺d 2 ⬇ ⫺0.017 n共n2 ⫺ 1兲
Fail to reject H0. There is not enough evidence to support the claim. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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11.5 THE RUNS TEST
11.5 Try It Yourself Solutions 1a. P P P F P F P P P P F F P F P P F F F P P P F P P P b. 13 groups ⇒ 13 runs c. 3, 1, 1, 1, 4, 2, 1, 1, 2, 3, 3, 1, 3 2a. H0 : The sequence of genders is random. Ha : The sequence of genders is not random. (claim) b. ␣ ⫽ 0.05 c. F F F M M F F M F M M F F F n1 ⫽ number of Fs ⫽ 9 n2 ⫽ number of Ms ⫽ 5 G ⫽ number of runs ⫽ 7 d. cv ⫽ 3 and 12 e. G ⫽ 7 f. Fail to reject H0 . g. At the 5% significance level, there is not enough evidence to support the claim that the sequence of genders is not random. 3a. H0 : The sequence of weather conditions is random. Ha : The sequence of weather conditions is not random. (claim) b. ␣ ⫽ 0.05 c. n1 ⫽ number of Ns ⫽ 21 n2 ⫽ number of Ss ⫽ 10 G ⫽ number of runs ⫽ 17 d. cv ⫽ ± 1.96 e. G ⫽
G ⫽
2n1n2 2共21兲共10兲 ⫹1⫽ ⫹ 1 ⬇ 14.5 n1 ⫹ n2 21 ⫹ 10
冪共2nn n⫹共n2n兲 n共n ⫺⫹nn ⫺⫺n1兲兲 ⫽ 冪2共21共21兲共10⫹兲共102共21兲 共兲共2110⫹兲 ⫺1021⫺⫺1兲10兲 ⬇ 2.4 1 2
1
z⫽
2
1 2 2
1
1
2
2
2
G ⫺ G 17 ⫺ 14.5 ⫽ ⫽ 1.04 G 2.4
f. Fail to reject H0 . g. At the 5% significance level, there is not enough evidence to support the claim that the sequence of weather conditions each day is not random.
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CHAPTER 11

NONPARAMETRIC TESTS
11.5 EXERCISE SOLUTIONS 1. Number of runs ⫽ 8 Run lengths ⫽ 1, 1, 1, 1, 3, 3, 1, 1
2. Number of runs ⫽ 9 Run lengths ⫽ 2, 2, 1, 1, 2, 2, 1, 1, 2
3. Number of runs ⫽ 9 Run lengths ⫽ 1, 1, 1, 1, 1, 6, 3, 2, 4
4. Number of runs ⫽ 10 Run lengths ⫽ 3, 3, 1, 2, 6, 1, 2, 1, 1, 2
5. n1 ⫽ number of Ts ⫽ 6 n2 ⫽ number of Fs ⫽ 6
6. n1 ⫽ number of U s ⫽ 8 n2 ⫽ number of Ds ⫽ 6
7. n1 ⫽ number of Ms ⫽ 10 n2 ⫽ number of Fs ⫽ 10
8. n1 ⫽ number of A s ⫽ 13 n2 ⫽ number of Bs ⫽ 9
9. n1 ⫽ number of Ts ⫽ 6 n1 ⫽ number of Fs ⫽ 6 cv ⫽ 3 and 11 11. n1 ⫽ number of Ns ⫽ 11 n1 ⫽ number of Ss ⫽ 7 cv ⫽ 5 and 14
10. n1 ⫽ number of Ms ⫽ 9 n2 ⫽ number of Fs ⫽ 3 cv ⫽ 2 and 8 12. n1 ⫽ number of Xs ⫽ 7 n2 ⫽ number of Ys ⫽ 14 cv ⫽ 5 and 15
13. (a) H0: The coin tosses were random. Ha: The coin tosses were not random. (claim) (b) n1 ⫽ number of Hs ⫽ 7 n2 ⫽ number of Ts ⫽ 9 cv ⫽ 4 and 14 (c) G ⫽ 9 runs (d) Fail to reject H0. (e) At the 5% significance level, there is not enough evidence to support the claim that the coin tosses were not random. 14. (a) H0: The selection of members was random. Ha: The selection of members was not random. (claim) (b) n1 ⫽ number of Rs ⫽ 18 n2 ⫽ number of Ds ⫽ 16 cv ⫽ 11 and 25 (c) G ⫽ 19 runs (d) Fail to reject H0. (e) At the 5% significance level, there is not enough evidence to support the claim that the selection of members was not random.
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15. (a) H0: The sequence of digits was randomly generated. Ha: The sequence of digits was not randomly generated. (claim) (b) n1 ⫽ number of Os ⫽ 16 n2 ⫽ number of Es ⫽ 16 cv ⫽ 11 and 23 (c) G ⫽ 9 runs (d) Reject H0. (e) At the 5% significance level, there is enough evidence to support the claim that the sequence of digits was not randomly generated. 16. (a) H0: The microchips are random by gender. (claim) Ha: The microchips are not random by gender. (b) n1 ⫽ number of Ms ⫽ 9 n2 ⫽ number of Fs ⫽ 20 cv ⫽ 8 and 18 (c) G ⫽ 12 runs (d) Fail to reject H0. (e) At the 5% significance level, there is not enough evidence to reject the claim that the microchips are random by gender. 17. (a) H0: The sequence is random. Ha: The sequence is not random. (claim) (b) n1 ⫽ number of Ns ⫽ 40 n2 ⫽ number of Ps ⫽ 9 cv ⫽ ± 1.96 (c) G ⫽ 14 runs 2n1n2 2共40兲共9兲 G ⫽ ⫹1⫽ ⫹ 1 ⬇ 15.7 n1 ⫹ n2 40 ⫹ 9
G ⫽
冪共2nn n⫹共n2n兲 n共n ⫺⫹nn ⫺⫺n1兲兲 ⫽ 冪2共40共40兲共9⫹兲共29共兲40共40兲共9兲⫹⫺9 40⫺ ⫺1兲 9兲 ⫽ 2.05 1 2
1
z⫽
2
1 2 2
1
1
2
2
2
G ⫺ G 14 ⫺ 15.7 ⫽ ⫽ ⫺0.83 G 2.05
(d) Fail to reject H0. (e) At the 5% significance level, there is not enough evidence to support the claim that the sequence is not random.
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CHAPTER 11

NONPARAMETRIC TESTS
18. (a) H0: The sequence of past winners is random. Ha: The sequence of past winners is not random. (claim) (b) n1 ⫽ number of Ts ⫽ 42 n2 ⫽ number of Ls ⫽ 16 cv ⫽ ± 1.96 (c) G ⫽ 32 runs
G ⫽
2n1n2 2共42兲共16兲 ⫹1⫽ ⫹ 1 ⫽ 24.2 n1 ⫹ n2 42 ⫹ 16
G ⫽
冪共2nn n⫹共n2n兲 n共n ⫺⫹nn ⫺⫺n1兲兲 ⫽ 冪2共42共42兲共16⫹兲共162共42兲 共兲共4216⫹兲 ⫺1642⫺⫺1兲16兲 ⫽ 3.0 1 2
1
z⫽
2
1 2 2
1
1
2
2
2
G ⫺ G 32 ⫺ 24.2 ⫽ ⫽ 2.6 G 3.0
(d) Reject H0. (e) At the 5% significance level, there is enough evidence to support the claim that the sequence is not random. 19. H0: Daily high temperatures occur randomly. Ha: Daily high temperatures do not occur randomly. (claim) median ⫽ 87 n1 ⫽ number above median ⫽ 14 n2 ⫽ number below median ⫽ 13 cv ⫽ 9 and 20 G ⫽ 11 runs Fail to reject H0. At the 5% significance level, there is not enough evidence to support the claim that the daily high temperatures do not occur randomly. 20. H0: The selection of students’ GPAs was random. Ha: The selection of students’ GPAs was not random. (claim) median ⫽ 1.8 n1 ⫽ number above median ⫽ 12 n2 ⫽ number below median ⫽ 3 cv ⫽ 2 and 8 G ⫽ 3 runs Fail to reject H0. At the 5% significance level, there is not enough evidence to support the claim that the selection of students’ GPAs was not random. 21. Answers will vary.
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CHAPTER 11 REVIEW EXERCISE SOLUTIONS 1. (a) H0: median ⫽ $24,300 (claim); Ha: median ⫽ $24,300 (b) Critical value is 2. (c) x ⫽ 7 (d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to reject the claim that the median value of stock among families that own stock is $24,300. 2. (a) H0: median ⱕ $2000; Ha : median > $2000 (claim) (b) The critical value is 1. (c) x ⫽ 6 (d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that the median credit card debt is more than $2000. 3. (a) H0: median ⱕ 6 (claim); Ha: median > 6 (b) Critical value is z0 ⬇ ⫺1.28. (c) x ⫽ 44 共x ⫺ 0.5兲 ⫺ 0.5共n兲 共44 ⫺ 0.5兲 ⫺ 0.5共70兲 8.5 z⫽ ⫽ ⫽ ⬇ 2.03 4.1833 冪n 冪70 2 2 (d) Reject H0. (e) There is enough evidence at the 10% significance level to reject the claim that the median turnover time is no more than 6 hours. 4. (a) H0: There was no reduction in diastolic blood pressure. (claim) Ha: There was a reduction in diastolic blood pressure. (b) The critical value is 1. (c) x ⫽ 4 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to reject the claim that there was no reduction in diastolic blood pressure. 5. (a) H0: There is no reduction in diastolic blood pressure. (claim) Ha: There is a reduction in diastolic blood pressure. (b) Critical value is 2. (c) x ⫽ 3 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to reject the claim that there was no reduction diastolic blood pressure.
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CHAPTER 11

NONPARAMETRIC TESTS
6. (a) H0: median ⫽ $40,200 (claim); Ha: median ⫽ $40,200 (b) The critical values are z0 ⫽ ± 1.96. (c) x ⫽ 21
共x ⫹ 0.5兲 ⫺ 0.5共n兲 共21 ⫹ 0.5兲 ⫺ 0.5共54兲 ⫺5.5 ⫽ ⫽ ⬇ ⫺1.50 3.674 冪n 冪54 2 2 (d) Fail to reject H0. z⫽
(e) There is not enough evidence at the 5% significance level to reject the claim that the median starting salary is $40,200. 7. (a) Independent; Wilcoxon Rank Sum Test (b) H0: There is no difference in the amount of time that it takes to earn a doctorate. Ha: There is a difference in the amount of time that it takes to earn a doctorate. (claim) (c) Critical values are z0 ⫽ ± 2.575. (d) R ⫽ 173.5 n 共n ⫹ n2 ⫹ 1兲 12共12 ⫹ 12 ⫹ 1兲 R ⫽ 1 1 ⫽ ⫽ 150 2 2
R ⫽ z⫽
冪n n 共n 12⫹ n 1 2
1
2
⫹ 1兲
冪共12兲共12兲共1212⫹ 12 ⫹ 1兲 ⬇ 17.321
R ⫺ R 173.5 ⫺ 150 ⫽ ⬇ 1.357 R 17.321
(e) Fail to reject H0. (f) There is not enough evidence at the 1% significance level to support the claim that there is a difference in the amount of time that it takes to earn a doctorate. 8. (a) Dependent; Wilcoxon SignedRank Test (b) H0: The new drug does not affect the number of headache hours experienced. Ha: The new drug does affect the number of headache hours experienced. (claim) (c) The critical value is 4. (d) ws ⫽ 1.5 (e) Reject H0. (f) There is enough evidence at the 5% significance level to support the claim that the new drug does affect the number of headache hours experienced. 9. (a) H0: There is no difference in salaries between the fields of study. Ha: There is a difference in salaries between the fields of study. (claim) (b) Critical value is 5.991. (c) H ⬇ 21.695 (d) Reject H0. (e) There is enough evidence at the 5% significance level to conclude that there is a difference in salaries between the fields of study. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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10. (a) H0: There is no difference in the amount of time to earn a doctorate between the fields of study. Ha: There is a difference in the amount of time to earn a doctorate between the fields of study. (claim) (b) The critical value is 5.991. (c) H ⬇ 3.282 (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to support the claim that there is a difference in the amount of time to earn a doctorate between the fields of study. 11. (a) H0: s ⫽ 0; Ha: s ⫽ 0 (claim) (b) Critical value is 0.881. (c) 兺d 2 ⫽ 57 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.321 n共n2 ⫺ 1兲
(d) Fail to reject H0. (e) There is not enough evidence at the 1% significance level to support the claim that there is a correlation between overall score and price. 12. (a) H0: s ⫽ 0; Ha: s ⫽ 0 (claim) (b) The critical value is 0.700. (c) 兺d 2 ⫽ 94.5 rs ⫽ 1 ⫺
6兺d 2 ⬇ 0.213 n共n2 ⫺ 1兲
(d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to support the claim that there is a correlation between overall score and price. 13. (a) H0: The traffic stops were random by gender. Ha: The traffic stops were not random by gender. (claim) (b) n1 ⫽ number of Fs ⫽ 12 n2 ⫽ number of Ms ⫽ 13 cv ⫽ 8 and 19 (c) G ⫽ 14 runs (d) Fail to reject H0. (e) There is not enough evidence at the 5% significance level to support the claim that the traffic stops were not random.
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CHAPTER 11

NONPARAMETRIC TESTS
14. (a) H0: The departure status of buses is random. Ha: The departure status of buses is not random. (claim) (b) n1 ⫽ number of Ts ⫽ 11 n2 ⫽ number of Ls ⫽ 7 cv ⫽ 5 and 14 (c) G ⫽ 5 runs (d) Reject H0. (e) There is enough evidence at the 5% significance level to support the claim that the departure status of buses is not random.
CHAPTER 11 QUIZ SOLUTIONS 1. (a) H0: There is no difference in the salaries between genders. Ha: There is a difference in the salaries between genders. (claim) (b) Wilcoxon Rank Sum Test (c) Critical values are z0 ⫽ ± 1.645. (d) R ⫽ 67.5 n 共n ⫹ n2 ⫹ 1兲 10共10 ⫹ 10 ⫹ 1兲 R ⫽ 1 1 ⫽ ⫽ 105 2 2
R ⫽ z⫽
冪n n 共n 12⫹ n 1 2
1
2
⫹ 1兲
⫽
冪共10兲共10兲共1012⫹ 10 ⫹ 1兲 ⬇ 13.229
R ⫺ R 67.5 ⫺ 105 ⫽ ⬇ ⫺2.835 R 13.229
(e) Reject H0. (f) There is enough evidence at the 10% significance level to support the claim that there is a difference in the salaries between genders. 2. (a) H0: median ⫽ 50 (claim); Ha: median ⫽ 50 (b) Sign Test
(c) Critical value is 5.
(d) x ⫽ 9
(e) Fail to reject H0.
(f) There is not enough evidence at the 5% significance level to reject the claim that the median number of annual volunteer hours is 50. 3. (a) H0: There is no difference in rent between regions. Ha: There is a difference in rent between regions. (claim) (b) KruskalWallis Test
(c) Critical value is 7.815.
(d) H ⬇ 11.826
(e) Reject H0.
(f) There is enough evidence at the 5% significance level to conclude that there is a difference in rent between regions.
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4. (a) H0: The days with rain are random. Ha: The days with rain are not random. (claim) (b) The Runs test (c) n1 ⫽ number of N s ⫽ 15 n2 ⫽ number of Rs ⫽ 15 cv ⫽ 10 and 22 (d) G ⫽ 16 runs (e) Fail to reject H0. (f) There is not enough evidence at the 5% significance level to conclude that days with rain are not random.
CUMULATIVE REVIEW FOR CHAPTERS 9–11 r ⫽ 0.828
y
Women’s time (in seconds)
1. (a)
12.5
There is a strong positive linear correlation.
12.0 11.5 11.0 10.5 x 9.5
10
10.5
11
Men’s time (in seconds)
(b) H0: ⫽ 0 Ha: ⫽ 0 (claim) t ⫽ 5.911 p ⫽ 0.0000591 Reject H0. There is enough evidence at the 5% significance level to conclude that there is a significant linear correlation. (c) y ⫽ 1.423x ⫺ 3.204 ^
Women’s time (in seconds)
y 12.5 12.0 11.5 11.0 10.5 x 9.5
10
10.5
11
Men’s time (in seconds)
(d) y ⫽ 1.423共9.90兲 ⫺ 3.204 ⫽ 10.884 seconds ^
2. H0: Median 共Union兲 ⫽ Median 共Non union兲 H1: Median 共Union兲 ⫽ Median 共Non union兲 (claim) cv ⫽ ± 1.96 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 11

NONPARAMETRIC TESTS
R ⫽ 145.0
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 10共10 ⫹ 10 ⫹ 1兲 ⫽ ⫽ 105 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪10共10兲共1012⫹ 10 ⫹ 1兲 ⫽ 13,229
R ⫺ R 145.0 ⫺ 105 ⫽ ⫽ 4.375 R 13,229
Reject H0. There is enough evidence at the 5% significance level to support the claim. 3. H0: median ⫽ 48 (claim) H1: median ⫽ 48 cv ⫽ 3 x⫽7 Fail to reject H0. There is not enough evidence at the 5% significance level to reject the claim. 4. H0: 1 ⫽ 2 ⫽ 3 ⫽ 4 (claim) H1: At least one if different. cv ⫽ 2.29 F ⫽ 2.476 Reject H0. There is enough evidence at the 10% significance level to reject the claim. 5. H0: 21 ⫽ 22 (claim) H1: 21 ⫽ 22 cv ⫽ 2.01 F⫽
s21 34.6 ⫽ ⫽ 1.042 s22 33.2
Fail to reject H0. There is not enough evidence at the 10% significance level to reject the claim. 6. H0: The medians are all equal. H1: The medians are not all equal. (claim) cv ⫽ 11.345 H ⫽ 14.78 Reject H0. There is enough evidence at the 1% significance level to support the claim. 7.
Physicians
Distribution
1 2–4 5–9 10+
0.36 0.32 0.20 0.12
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H0: The distribution is as claimed. (claim) H1: The distribution is not as claimed. Physician
Distribution
Observed
Expected
1 2–4 5–9 10+
0.36 0.32 0.20 0.12
116 84 66 23
104.04 92.48 57.80 34.68
冇O ⴚ E冈2 E 1.375 0.778 1.163 3.934 x 2 ⫽ 7.250
289
cv ⫽ 7.815 x2 ⫽ 7.250 Fail to reject H0. There is not enough evidence at the 5% significance level to reject the claim. 8. (a) r 2 ⫽ 0.733 Metacarpal bone length explains 73.3% of the variability in height. About 26.7% of the variation is unexplained. (b) se ⫽ 4.255 (c) y ⫽ 94.428 ⫹ 1.700共50兲 ⫽ 179.428 ^
冪1 ⫹ n1 ⫹ n兺xn共x ⫺⫺共x兺x兲 兲
E ⫽ t c Se
2
0 2
2
共50 ⫺ 45,444兲 冪1 ⫹ 19 ⫹ 9共918,707 兲 ⫺ 共409兲
⫽ 2.365共4.255兲
⫽ 2.365共4.255兲冪1.284 ⫽ 11.402
2
2
y ± E ⇒ 共168.026, 190.83兲 ^
You can be 95% confident that the height will be between 168.026 centimeters and 190.83 centimeters when the metacarpal bone length is 50 centimeters. 9.
Score
Rank
Price
Rank
d
d2
85 83 83 81 77 72 70 66
8 6.5 6.5 5 4 3 2 1
81 78 56 77 62 85 62 61
7 6 1 5 3.5 8 3.5 2
1.0 0.5 5.5 0 0.5 ⫺5.0 ⫺1.5 ⫺1.0
1.00 0.25 30.25 0.00 0.25 25.00 2.25 1.00 兺d 2 ⫽ 60
H0: s ⫽ 0 H1: s ⫽ 0 (claim) cv ⫽ 0.643 rs ⫽ 1 ⫺
6兺d 2 60共60兲 ⫽1⫺ ⫽ 0.286 2 n共n ⫺ 1兲 8共82 ⫺ 1兲
Fail to reject H0. There is not enough evidence at the 10% significance level to support the claim. 10. (a) y ⫽ 91.113 ⫺ 0.014共4325兲 ⫹ 0.018共1900兲 ⫽ 64.763 bushels ^
(b) y ⫽ 91.113 ⫺ 0.014共4900兲 ⫹ 0.018共2163兲 ⫽ 61.447 bushels ^
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Alternative Presentation of the Standard Normal Distribution
APPENDIX
A
Try It Yourself Solutions 1 (1) 0.4857 (2) z ⫽ ± 2.17 2a.
z 0
2.13
b. 0.4834 c. Area ⫽ 0.5 ⫹ 0.4834 ⫽ 0.9834 3a.
z −2.16
0
b. 0.4846 c. Area ⫽ 0.5 ⫹ 0.4846 ⫽ 0.9846 4a.
z −2.16
−1.35 0
b. z ⫽ ⫺2.16: Area ⫽ 0.4846 z ⫽ ⫺1.35: Area ⫽ 0.4115 c. Area ⫽ 0.4846 ⫺ 0.4115 ⫽ 0.0731
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Normal Probability Plots and Their Graphs
APPENDIX
C
Try It Yourself Solutions 3
1a. 8000
42,000
−3
The points do not appear to be approximately linear. b. 39,860 is a possible outlier because it is far removed from the other entries in the data set. c. Because the points do not appear to be approximately linear and there is an outlier, you can conclude that the sample data do not come from a population that has a normal distribution.
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ACTIVITIES
ACTIVITY 1.3 1. 再1, 17, 17, 17, 14, 10, 15, 6冎 Answers will vary. This list is a random sample taken with replacement because 17 appears 3 times. 2. Min ⫽ 1 Max ⫽ 731 Number of samples ⫽ 8
再565, 718, 305, 75, 364, 221, 230, 231冎 Answers will vary. The random number generator is easier to use than a random number table.
ACTIVITY 2.3 1. When the mean is equal to the median, the shape of the distribution will be symmetric. When a few points are added below 10, the mean shifts downward. As points are added below 10, both the mean and median will shift downward. 2. No, the mean and median cannot be any of the points that were plotted.
ACTIVITY 2.4 1. 再18, 10, 15, 20, 16, 14, 19, 17, 13, 15冎 x ⫽ 15.7 s ⫽ 2.98 By adding x ⫽ 15, x ⫽ 15.6 and s ⫽ 2.84, the mean moves slightly closer to 15 while the standard deviation decreases. By adding x ⫽ 20, x ⫽ 16 and s ⫽ 2.98, the mean and standard deviation increase. 2. 再30, 30, 30, 30, 40, 40, 40, 40冎 s ⫽ 5.35
再35, 35, 35, 35, 35, 35, 35, 35冎 s⫽0 When the values in a data set are all equal, the standard deviation will be zero.
ACTIVITY 3.1 1. Answers will vary. 2. P共market goes up on day 36兲 ⫽ 0.5
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ACTIVITIES
ACTIVITY 3.3 1. P共3 or 4兲 ⫽ 16 ⫹ 16 ⫽ 13 2. Answers will vary. 3. The green line will increase to 12.
ACTIVITY 4.2 1. 再7, 8, 7, 7, 7, 7, 8, 8, 9, 10冎 Answers will vary. 0 (a) P共x ⫽ 5兲 ⫽ 10 5 (b) P共at least 8兲 ⫽ 10 5 (c) P共at most 7兲 ⫽ 10
2. 再3, 1, 2, 2, 1, 2, 2, 2, 5, 3冎 Answers will vary. 0 (a) P共x ⫽ 4兲 ⫽ 10 1 (b) P共at least 5兲 ⫽ 10 9 (c) P共less than 4兲 ⫽ 10
3. Answers will vary. P共x ⫽ 5兲 ⬇ 0.103 When using N ⫽ 100, the estimated probability of exactly five is closer to 0.103.
ACTIVITY 5.4 1. The mean of the sampling distribution of a uniform, bellshaped, and skewed distribution will be approximately 25. 2. The estimated standard deviation of the sampling distribution will be approximately
冪50
.
ACTIVITY 6.2 1. Approximately 95% of the z and t confidence intervals will contain the mean of 25. 2. Approximately 95% of the z and t confidence intervals will contain the mean of 7.26. Because n ⫽ 24, we should use a tCI.
ACTIVITY 6.3 1. Approximately 95% and 99% of the CI’s will contain 0.6. 2. Approximately 95% and 99% of the CI’s will contain 0.4.
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ACTIVITIES
ACTIVITY 7.2 1. At ␣ ⫽ 0.05, approximately 50 of the 1000 null hypotheses will be rejected. At ␣ ⫽ 0.01, approximately 10 of the 1000 null hypotheses will be rejected. 2. If the null hypothesis is rejected at the 0.01 level, the pvalue will be smaller than 0.01. So, it would also be rejected at the 0.05 level. Suppose a null hypothesis is rejected at the 0.05 level. It will not necessarily be rejected at the 0.01 level because 0.01 < pvalue < 0.05. 3. H0: ⱖ 27 The proportion of rejected null hypotheses will be larger than 0.05 and 0.01 because the true mean is 25 共i.e. < 27兲.
ACTIVITY 7.4 1. Approximately 50 null hypotheses will be rejected at the 0.05 level while approximately 10 null hypotheses will be rejected at the 0.01 level. 2. H0: p ⱖ 0.4 The proportion of null hypotheses rejected will be less than 0.05 and 0.01 because the true p ⫽ 0.6.
ACTIVITY 9.1 1. If the scatter plot is linear with a positive slope, r ⬇ 1. If the scatter plot is linear with a negative slope r ⬇ ⫺1. 2. Create a data set that is linear with a positive slope. 3. Create a data set that is nonlinear. 4. r ⬇ ⫺0.9 has a negative slope while using r ⬇ 0.9 will have a positive slope.
ACTIVITY 9.2 1. Answers will vary. 2. The regression line is influenced by the new point. 3. The regression line is not as influenced by the new point due to more points being used to calculate the regression line. 4. As the sample size increases, the slope changes less.
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355

CASE STUDY
CHAPTER 1
CASE STUDY: RATING TELEVISION SHOWS IN THE UNITED STATES 1. Yes. A rating of 8.4 is equivalent to 9,357,600 households which is twice the number of households at a rating of 4.2. 2.
10,000 ⬇ 0.00897 ⇒ 0.897% 1,114,000
3. Program Name and Network 4. Rank and Rank Last Week Data can be arranged in increasing or decreasing order. 5. Day, Time Data can be chronologically ordered. Hours or minutes 6. Rating, Share, and Audience 7. Shows are ranked by rating. Share is not arranged in decreasing order. 8. A decision of whether a program should be cancelled or not can be made based on the Nielsen ratings.
CHAPTER 2
CASE STUDY: EARNINGS OF ATHLETES 1. NFL has $3, 041,000,000 in total revenue. 2. MLB: ⫽ $3,152,305.83
MLS: ⫽ $265,576.32
NBA: ⫽ $4,414,977.48
NFL: ⫽ $1,620,138.52
NHL: ⫽ $1,960,454.32
NASCAR: ⫽ $2,500,000.00
PGA: ⫽ $1,167,300.38 3. NBA has a mean player revenue of $4,414,977. 4. MLB: s ⫽ $4,380,876.18
MLS: s ⫽ $124,022.66
NBA: s ⫽ $4,533,634.75
NFL: s ⫽ $2,220,852.88
NHL: s ⫽ $1,588,168.87
NASCAR: S ⫽ $2,228,796.19
PGA: $1,271,590.31 5. NBA 6. By examining the frequency distributions, the NBA is more bellshaped.
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CASE STUDY
CHAPTER 3
CASE STUDY: PROBABILITY AND PARKING LOT STRATEGIES 1. No, each parking space is not equally likely to be empty. Spaces near the entrances are more likely to be filled. 2. PickaLane strategy appears to take less time (in both time to find a parking space and total time in parking lot). 3. Cycling strategy appears to give a shorter average walking distance to the store. 4. In order to assume that each space is equally likely to be empty, drivers must be able to see which spaces are available as soon as they enter a row. 5. P共1st row兲 ⫽
1 since there are seven rows. 7
6. P共in one of 20 closest spaces兲 ⫽
20 72
7. Answers will vary. Using the PickaLane strategy, the probability that you can find a parking space in the most desirable category is low. Using the Cycling strategy, the probability that you can find a parking space in the most desirable category depends on how long you cycle. The longer you cycle, the probability that you can find a parking space in the most desirable category will increase.
CHAPTER 4
CASE STUDY: BINOMIAL DISTRIBUTION OF AIRPLANE ACCIDENTS 1. P共crash in 2006兲 ⫽
2 ⫽ 0.00000018 11,000,000
2. (a) P共4兲 ⫽ 0.192 (b) P共10兲 ⫽ 0.009 (c) P共1 ⱕ x ⱕ 5兲 ⫽ P共1兲 ⫹ P共2兲 ⫹ P共3兲 ⫹ P共4兲 ⫹ P共5兲 ⫽ 0.054 ⫹ 0.119 ⫹ 0.174 ⫹ 0.192 ⫹ 0.169 ⫽ 0.708 3. n ⫽ 11,000,000, p ⫽ 0.0000008 (a) P共4兲 ⬇ 0.0376641 (b) P共10兲 ⬇ 0.1156838 (c) P共1 ⱕ x ⱕ 5兲 ⬇ 0.12823581
x
P冇 x冈
0 1 2 3 4 5 6 7 8 9 10 11 12
0.0001507 0.0013264 0.0058364 0.0171200 0.0376641 0.0662889 0.0972237 0.1222241 0.1344465 0.1314588 0.1156838 0.0925470 0.0678678
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CASE STUDY
4. No, because each flight is not an independent trial. There are many flights that use the same plane. 5. Assume p ⫽ 0.0000004. The number of flights would be 64,000 years ⭈ 365 days per year ⫽ 23,360,000 flights. The average number of fatal accidents for that number of flights would be 9.344. So, the claim cannot be justified.
CHAPTER 5
CASE STUDY: BIRTH WEIGHTS IN AMERICA 1. (a) 40 weeks 共 ⫽ 7.69兲 (b) 32 to 35 weeks 共 ⫽ 5.71兲 (c) 41 weeks 共 ⫽ 7.79兲 2. (a) P共x < 5.5兲 ⫽ P共z < 3.02兲 ⫽ 0.9987 → 99.87% (b) P共x < 5.5兲 ⫽ P共z < ⫺0.14兲 ⫽ 0.4432 → 44.32% (c) P共x < 5.5兲 ⫽ P共z < ⫺1.66兲 ⫽ 0.0484 → 4.84% (d) P共x < 5.5兲 ⫽ P共z < ⫺1.90兲 ⫽ 0.0287 → 2.87% 3. (a) top 10% → 90th percentile → z ⬇ 1.285 x ⫽ ⫹ z ⫽ 7.31 ⫹ 共1.285兲共1.09兲 ⬇ 8.711 Birth weights must be at least 8.711 pounds to be in the top 10%. (b) top 10% → 90th percentile → z ⬇ 1.285 x ⫽ ⫹ z ⫽ 7.61 ⫹ 共1.285兲共1.12兲 ⬇ 9.036 Birth weights must be at least 9.036 pounds to be in the top 10%. 4. (a) P共6 < x < 9兲 ⫽ P共0.20 < z < 2.24兲 ⫽ 0.9875 ⫺ 0.5793 ⫽ 0.4082 (b) P共6 < x < 9兲 ⫽ P共⫺1.20 < z < 1.55兲 ⫽ 0.9394 ⫺ 0.1151 ⫽ 0.8243 (c) P共6 < x < 9兲 ⫽ P共⫺1.45 < z < 1.25兲 ⫽ 0.8944 ⫺ 0.0735 ⫽ 0.8209 5. (a) P共x < 3.3兲 ⫽ P共z < 1.18兲 ⫽ 0.8810 (b) P共x < 3.3兲 ⫽ P共z < ⫺1.64兲 ⫽ 0.0505 (c) P共x < 3.3兲 ⫽ P共z < ⫺3.68兲 ⫽ 0.001 (using technology)
CHAPTER 6
CASE STUDY: SHOULDER HEIGHTS OF APPALACHIAN BLACK BEARS 1. (a) x ⫽ 79.7 (b) x ⫽ 75.7
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360

CASE STUDY
2. (a) s ⫽ 12.1 (b) s ⫽ 7.6 3. (a) x ± zc
s
(b) x ± tc 4. x ± zc
冪n
(b) n ⫽
⇒ 79.7 ± 1.96
12.1 冪40
7.6
⇒ 75.7 ± 2.048
冪n
s
5. (a) n ⫽
s 冪n
⇒ 78.1 ± 1.96
冪28
2
zc ⌭
2
冢 冣 冢 ⫽
⫽ 共72.7, 78.6兲
10.6 ⫽ 共75.5, 80.6兲 冪68
冢 ⌭ 冣 ⫽ 冢2.5750.5⭈ 12.4冣 zc
⫽ 共76.0, 83.5兲
2.575 ⭈ 7.8 0.5
冣
2
2
⫽ 4078.10 ⇒ 4079 bears
⫽ 1613.63 ⇒ 1614 bears
CHAPTER 7
CASE STUDY: HUMAN BODY TEMPERATURE: WHAT’S NORMAL? 1. (a) (see part d) (b) z0 ⫽ ± 1.96 (see part d) (c) Rejection regions: z < ⫺1.96 and z > 1.96 (see part d) (d) x ⬇ 98.25, s ⬇ 0.73 z⫽
x ⫺ 98.25 ⫺ 98.6 ⫺0.35 ⫽ ⫽ ⬇ ⫺5.469 s 0.73 0.0640 冪n 冪130
1 α 2
1 α 2
= 0.025
−5 −4 −3 −2 −1
z = −5.469
−z0
= 0.025
z 0
1
2
3
4
5
z0
(e) Reject H0. (f) There is sufficient evidence at the 5% level to reject the claim that the mean body temperature of adult humans is 98.6⬚ F. 2. No, ␣ ⫽ 0.01 → z0 ⫽ ± 2.575 → Reject H0
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CASE STUDY
3. H0 : m ⫽ 98.6 and Ha: m ⫽ 98.6
␣ ⫽ 0.01 → z0 ⫽ ± 2.575 x ⬇ 98.105, z⫽
s ⬇ 0.699
x ⫺ 98.105 ⫺ 98.6 ⫺0.495 ⫽ ⫽ ⬇ ⫺5.709 s 0.699 0.0867 冪n 冪65
Reject H0. There is sufficient evidence to reject the claim. 4. H0: w ⫽ 98.6 and Ha: w ⫽ 98.6
␣ ⫽ 0.01 → z0 ⫽ ± 2.575 x ⬇ 98.394, z⫽
s ⬇ 0.743
x ⫺ 98.394 ⫺ 98.6 ⫺0.206 ⫽ ⫽ ⬇ ⫺2.234 s 0.743 0.0922 冪n 冪65
Fail to reject H0. There is insufficient evidence to reject the claim. 5. x ⬇ 98.25, x ± zc
s 冪n
s ⬇ 0.73 ⫽ 98.25 ± 2.575
0.73
98.25 ± 0.165 ⬇ 共98.05, 98.415兲
冪130
6. Wunderlich may have sampled more women than men, thus causing an overestimated body temperature.
CHAPTER 8
CASE STUDY: DASH DIET AND BLOOD PRESSURE 1. H0: 1 ⫽ 2 vs H1: 1 ⫽ 2 (claim) z ⫽ ± 1.96 Reject H0 if z < ⫺1.96 or z > 1.96 z⫽
冇x1 ⫺ x2 冈 ⫺ 冇0 冈
冪n
s 21 1
⫹
s 22
⫽
n2
10.36 ⫺ 5.73
冈 冪冇15.79 77
2
⫹
冇12.50冈 79
2
⫽
4.63 2.03 冪5.216
Reject H0. There is enough evidence to support the claim. 2. H0: 3 ⫽ 4 vs H2: 3 ⫽ 4 (claim) z0 ⫽ ± 1.96 Reject H0 if z < ⫺1.96 or z > 1.96. z⫽
共x3 ⫺ x4 兲 ⫺ 共0 兲
冪
s 24
⫽
4.85 ⫺ 3.53
冪
2
2
⫽
1.32
⬇ 0.64 共13.73兲 共12兲 ⫹ ⫹ n3 n4 76 79 Fail to reject H0. There is not enough evidence to support the claim. s32
冪4.303
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CASE STUDY
3. H0: 2 ⫽ 4 vs H1: 2 ⫽ 4 (claim) z ⫽ ± 1.96 Reject H0 if z < ⫺1.96 or z > 1.96. z⫽
共x2 ⫺ x4 兲 ⫺ 共0 兲
冪n
⫽
5.73 ⫺ 3.53
兲 冪共12.50 79
⫽
2.2 ⫽ 1.13 冪3.801
共12.00兲 n4 79 2 Fail to reject H0. There is not enough evidence to support the claim. s 22
s 24
⫹
2
2
⫹
4. H0: 1 ⫽ 3 vs H1: 1 ⫽ 3 (claim) z0 ⫽ ± 1.96 Reject H0 if z < ⫺1.96 or z > 1.96. z⫽
共x1 ⫺ x3 兲 ⫺ 共0 兲
冪ns
⫽
10.36 ⫺ 4.85
兲 冪共15.79 77
⫽
5.51 ⫽ 2.30 冪5.718
共13.73兲2 76 1 Reject H0. There is enough evidence to support the claim. 2 1
s 23 n3
⫹
2
⫹
5. (a) H0: 1 ⫽ 5 vs H1: 1 ⫽ 5 (claim) z ⫽ ± 2.576 Reject H0 if z < ⫺2.576 or z > 2.576. z⫽
共x1 ⫺ x2 兲 ⫺ 共0 兲
冪n
s 21
⫹
1
s 25
⫽
n5
10.36 ⫺ 7.05
冪
共15.79兲 共7.28兲 ⫹ 77 42 2
2
⫽
3.31 冪4.500
⫽ 1.56
Fail to reject H0. There is not enough evidence to support the claim. (b) H0: 4 ⫽ 5 vs H1: 4 ⫽ 5 (claim) z ⫽ ± 2.576 Reject H0 if z < ⫺2.576 or z > 2.576. z⫽
共x4 ⫺ x2 兲 ⫺ 共0 兲
冪
s 25
⫽
3.53 ⫺ 7.05
冪
2
2
⫽
⫺3.52
⫽ ⫺2.00 共12兲 共7.28兲 ⫹ ⫹ n4 n5 79 42 Fail to reject H0. There is not enough evidence to support the claim. s 24
冪3.085
6. There was enough evidence to support the claim that the mean weight losses of the Atkins diet and the LEARN diet were different. There was enough evidence to support the claim that the mean weight losses of the Atkins diet and the Ornish diet were different.
CHAPTER 9
CASE STUDY: CORRELATION OF BODY MEASUREMENTS 1. Answers will vary.
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 2.
CASE STUDY
r a b c d e f g h i j k l
0.698 0.746 0.351 0.953 0.205 0.580 0.844 0.798 0.116 0.954 0.710 0.710
3. d, g, and j have strong correlations 共r > 0.8兲. (d) y ⫽ 1.129x ⫺ 8.817 ^
(g) y ⫽ 1.717x ⫹ 23.144 ^
(j) y ⫽ 1.279x ⫹ 10.451 ^
4. (a) y ⫽ 0.086共180兲 ⫹ 21.923 ⫽ 37.403 ^
(b) y ⫽ 1.026共100兲 ⫺ 15.323 ⫽ 87.277 ^
5. (weight, chest) → r ⫽ 0.888 r ⫽ 0.930 (weight, hip) r ⫽ 0.849 (neck, wrist) r ⫽ 0.953 (chest, hip) r ⫽ 0.936 (chest, thigh) r ⫽ 0.855 (chest, knee) r ⫽ 0.908 (chest, ankle) r ⫽ 0.863 (abdom, thigh) r ⫽ 0.903 (abdom, knee) r ⫽ 0.894 (hip, thigh) r ⫽ 0.908 (hip, ankle) r ⫽ 0.954 (thigh, knee) r ⫽ 0.874 (thigh, ankle) r ⫽ 0.857 (ankle, knee) r ⫽ 0.854 (forearm, wrist)
CHAPTER 10 CASE STUDY: TRAFFIC SAFETY FACTS 1. 43,443共0.24兲 ⬇ 10,426.32 ⬇ 10,427 2.
3478 ⬇ 0.243 > 0.24 → Central United States and 14312 1655 ⬇ 0.245 > 0.24 → Western United States 6751
3. The number of fatalities in the Eastern, Central, and Western United States for the 25–34 age group exceeded the expected number of fatalities.
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CASE STUDY
4. H0: Region is independent of age. Ha: Region is dependent on age. d.f. ⫽ 共r ⫺ 1兲共c ⫺ 1兲 ⫽ 14
20 ⫽ 23.685 → Reject H0 if 2 > 23.685. 2 ⬇ 71.504 Reject H0. There is enough evidence at the 5% significance level to conclude that region is dependent on age. 5. H0: Eastern United States has same distribution as the entire United States. Ha: Eastern United States has a different distribution than the entire United States.
20 ⫽ 14.067 → Reject H0 if 2 > 14.067. 2 ⬇ 14.694 Reject H0. There is enough evidence at the 5% significance level to conclude that the eastern United States has a different distribution than the entire United States. 6. H0: Central United States has same distribution as the entire United States. Ha: Central United States has a different distribution than the entire United States.
20 ⫽ 14.067 → Reject H0 if 2 > 14.067. 2 ⬇ 38.911 Reject H0. There is enough evidence at the 5% significance level to conclude that the central United States has a different distribution than the entire United States. 7. H0: Western United States has same distribution as the entire United States. Ha: Western United States has a different distribution than the entire United States.
20 ⫽ 14.067 → Reject H0 if 2 > 14.067. 2 ⬇ 58.980 Reject H0. There is enough evidence at the 5% significance level to conclude that the western United States has a different distribution than the entire United States. 8. Alcohol, vehicle speed, driving conditions, etc.
CHAPTER 11
CASE STUDY: EARNINGS BY COLLEGE DEGREE 1. (a), (b), (c), (d) The doctorate groups appear to make more money than the bachelor groups.
Earnings by Highest Degree (in dollars) All doctorates Male doctorates All Bachelors Female Bachelors
60,000
120,000
180,000
Earnings (in dollars)
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CASE STUDY
2. H0: median ⱕ $92,000 (claim); Ha: median > $92,000 The critical value is 1. x⫽5 Fail to reject H0. There is not enough evidence to reject the claim. 3. H0: median ⫽ $102,500 (claim); Ha: median ⫽ $102,500 The critical value is 1. x⫽3 Fail to reject H0. There is not enough evidence to reject the claim. 4. H0: median ⱖ $56,000 (claim); Ha: median < $56,000 The critical value is 1. x⫽4 Fail to reject H0. There is not enough evidence to reject the claim. 5. H0: median ⫽ $42,000; Ha: median ⫽ $42,000 (claim) The critical value is 1. x⫽1 Reject H0. There is enough evidence to support the claim. 6. H0: There is no difference in median earnings. (claim) Ha: There is a difference in median earnings. The critical values are z0 ⫽ ± 2.575. R ⫽ 98
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 10共10 ⫹ 10 ⫹ 1兲 ⫽ ⫽ 105 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪共10兲共10兲共1012⫹ 10 ⫹ 1兲 ⬇ 13.2
R ⫺ R 98 ⫺ 105 ⫽ ⬇ ⫺0.530 R 13.2
Fail to reject H0. There is not enough evidence to reject the claim. 7. H0: There is no difference in median earnings. Ha: There is a difference in median earnings. (claim) The critical values are z0 ⫽ ± 2.575. R ⫽ 140
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 10共10 ⫹ 10 ⫹ 1兲 ⫽ ⫽ 105 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪共10兲共10兲共1012⫹ 10 ⫹ 1兲 ⬇ 13.2
R ⫺ R 140 ⫺ 105 ⫽ ⬇ 2.652 R 13.2
Reject H0. There is enough evidence to support the claim.
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365

USES AND ABUSES
CHAPTER 1
USES AND ABUSES 1. Answers will vary. 2. Answers will vary.
CHAPTER 2
USES AND ABUSES 1. Answers will vary. 2. Answers will vary.
CHAPTER 3
USES AND ABUSES 1. (a) P共winning Tues and Wed兲 ⫽ P共winning Tues兲 ⭈ P共winning Wed兲 ⫽
1 1 冢1000 冣 ⭈ 冢1000 冣
⫽ 0.000001 (b) P共winning Wed given won Tues兲 ⫽ P共winning Wed兲 ⫽
1 1000
⫽ 0.001 (c) P共winning Wed given didn't win Tues兲 ⫽ P共winning Wed兲 ⫽
1 1000
⫽ 0.001 2. Answers will vary. P共 pickup or SUV兲 ⱕ 0.55 because P共pickup兲 ⫽ 0.25 and P共SUV兲 ⫽ 0.30, but a person may own both a pickup and an SUV (ie not mutually exclusive events). So, P共 pickup and SUV兲 ⱖ 0 and this probability would have to be subtracted from 0.55. If the events were mutually exclusive, the value of the probability would be 0.55. Otherwise, the probability will be less than 0.55 (not 0.60).
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367
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USES AND ABUSES
CHAPTER 4
USES AND ABUSES 1. 40 P共40兲 ⫽ 0.081 2. P共35 ⱕ x ⱕ 45兲 ⫽ P共35兲 ⫹ P共36兲 ⫹ . . . ⫹ P共45兲 ⫽ 0.7386 3. The probability of finding 36 adults out of 100 who prefer Brand A is 0.059. So the manufacturer’s claim is hard to believe. 4. The probability of finding 25 adults out of 100 who prefer Brand A is 0.000627. So the manufacturer’s claim is not be believable.
CHAPTER 5
USES AND ABUSES 1. ⫽ 100
⫽ 15 (a) z ⫽
x ⫺ 115 ⫺ 100 ⫽ ⫽ 1.73 兾冪n 15兾冪3
x ⫽ 115 is not an unusual sample mean. (b) z ⫽
x ⫺ 105 ⫺ 100 ⫽ ⫽ 1.49 兾冪n 15兾冪20
x ⫽ 105 is not an unusual sample mean. 2. The problem does not state the population of ages is normally distributed. 3. Answers will vary.
CHAPTER 6
USES AND ABUSES 1. Answers will vary. 2. Answers will vary.
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USES AND ABUSES
CHAPTER 7
USES AND ABUSES 1. Randomly sample car dealerships and gather data necessary to answer question. (Answers will vary.) 2. H0 : p ⫽ 0.57 We cannot prove that p ⫽ 0.57. We can only show that there is insufficient evidence in our sample to reject H0 : p ⫽ 0.57. (Answers will vary.) 3. If we gather sufficient evidence to reject the null hypothesis when it is really true, then a Type I error would occur. (Answers will vary.) 4. If we fail to gather sufficient evidence to reject the null hypothesis when it is really false, then a Type II error would occur. (Answers will vary.)
CHAPTER 8
USES AND ABUSES 1. Age and health. (Answers will vary.) 2. Blind: The patients do not know which group (medicine or placebo) they belong to. Double Blind: Both the researcher and patient do not know which group (medicine or placebo) that the patient belongs to. (Answers will vary.)
CHAPTER 9
USES AND ABUSES 1. Answers will vary. 2. Answers will vary. One example would be temperature and output of sulfuric acid. When sulfuric acid is manufactured, the amount of acid that is produced depends on the temperature at which the industrial process is run. As the temperature increases, so does the output of acid, up to a point. Once the temperature passes that point, the output begins to decrease. Amount of acid produced
y
x
Temperature
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369
370

USES AND ABUSES
CHAPTER 10
USES AND ABUSES 1. Answers will vary. An example would be a biologist who is analyzing effectiveness of three types of pesticides. The biologists sprays one pesticide on five different acres during week one. At the end of week one, he calculates the mean number of insects in the acres. He sprays the second pesticide on the same five acres during week two. At the end of week two, he calculates the mean number of insects in the acres. He follows a similar procedure with the third pesticide during week three. He wants to determine whether there is a difference in the mean number of insects per acre. Since the same five acres are treated each time, it is unclear as to the effectiveness of each of the pesticides. 2. Answers will vary. An example would be another biologist who is analyzing the effectiveness of three types of pesticides. He has fifteen acres to test. He randomly assigns each pesticide to five acres and treats all of the acres for three weeks. At the end of three weeks, the biologist calculates the mean number of insects for each of the pesticides. He wants to determine whether there is a difference in the mean number of insects per acre. Rejecting the null hypothesis would mean that at least one of the means differs from the others. In this example, rejecting the null hypothesis means that there is at least one pesticide whose mean number of insects per acre differs from the other pesticides.
CHAPTER 11
USES AND ABUSES 1. 共Answers will vary.兲 H0 : median ⱖ 10 H0 : median < 10 Using n ⫽ 5 and ␣ ⫽ 0.05, the cv ⫽ 0 共␣ ⫽ 0.031兲. Thus, every item in the sample would have to be less than 10 in order to reject the H0 . 共i.e. 100% of the sample would need to be less than 10.兲 However, using n ⫽ 20 and ␣ ⫽ 0.05, the cv ⫽ 6 共␣ ⫽ 0.057兲. Now only 14 items 共or 70%兲 of the sample would need to be less than 10 in order to reject H0 . 2. Nonparametric Test
Parametric Test
(a) Sign Test
ztest
(b) Paired Sample Sign Test
ttest
(c) Wilcoxon SignedRank Test
ttest
(d) Wilcoxon Rank Sum Test
ztest or ttest
(e) KruskalWallis Test
oneway Anova
(f) Spearmans Rank Correlation Coefficient
Pearson correlation coefficient
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REAL STATISTICS – REAL DECISIONS
CHAPTER 1
REAL STATISTICS–REAL DECISIONS 1. (a) If the survey identifies the type of reader that is responding, then stratified sampling ensures that representatives from each group of readers (engineers, manufacturers, researchers, and developers) are included in the sample. (b) Yes (c) After dividing the population of readers into their subgroups (strata), a simple random sample of readers of the same size is taken from each subgroup. (d) You may take too large a percentage of your sample from a subgroup of the population that is relatively small. 2. (a) Once the results have been compiled you have quantitative data, because “Percent Responding” is a variable consisting of numerical measurements. (b) “Percent Responding” is a ratio level of measurement since the data can be ordered and you can calculate meaningful differences between data entries. (c) Sample (d) Statistics 3. (a) Using Internet surveys introduces bias into the sampling process. The data was not collected randomly and may not adequately represent the population. (b) Stratified sampling would have been more appropriate assuming the population of readers consists of various subgroups that should be represented in the sample.
CHAPTER 2
REAL STATISTICS–REAL DECISIONS 1. (a) Examine data from the four cities and make comparisons using various statistical methodologies. For example, compare the average price of automobile insurance for each city. (b) Calculate: Mean, range, and population standard deviation for each city. 2. (a) Construct a Pareto chart because the data in use are quantitative and a Pareto chart positions data in order of decreasing height, with the tallest bar positioned at the left. (b)
City C
City B
City D
2200 2000 1800 1600 1400
City A
Price of insurance (in dollars)
Price of Insurance per City
City
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REAL STATISTICS – REAL DECISIONS
3. (a) Find the mean, range, and population standard deviation for each city. (b) City A: x ⫽ $2191.00
⬇ $351.86
range ⫽ $1015.00
City B: x ⫽ $2029.20
⬇ $437.54
range ⫽ $1336.00
City C: x ⫽ $1772.00
⬇ $418.52
range ⫽ $1347.00
City D: x ⫽ $1909.30
⬇ $361.14
range ⫽ $1125.00
(c) The mean from Cities B, C, and D appear to be similar. However, City A appears to have the highest price of insurance. These measurements support the conclusion in Exercise 2. 4. (a) You would tell your readers that on average, the price of automobile insurance is higher in this city than in other cities. (b) Location of city, weather, population.
CHAPTER 3
REAL STATISTICS–REAL DECISIONS 1. (a) Answers will vary. Investigate the probability of not matching any of the 5 white balls selected. (b) You could use the Multiplication Rule, the Fundamental Counting Principle, and Combinations. 1 2. If you played only the red ball, the probability of matching it is 42. However, because you must pick 5 white balls, you must get the white balls wrong. So, using the Multiplication Rule, we get
冢
冣
P matching only the red ball and not ⫽ P共matching red ball兲 ⭈ P共not matching any white balls兲 matching any of the 5 white balls ⫽
1 42
50
49
48
47
46
⭈ 55 ⭈ 54 ⭈ 53 ⭈ 52 ⭈ 51
⬇ 0.015 ⬇
3 . 200
3. Using combinations, calculate the number of ways to win, and divide by the number of outcomes.
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REAL STATISTICS – REAL DECISIONS
CHAPTER 4
REAL STATISTICS–REAL DECISIONS 1. (a) Answers will vary. For example, calculate the probability of obtaining zero clinical pregnancies out of 10 randomly selected ART cycles. (b) Binomial Discrete, we are counting the number of successes (clinical pregnancies). 2. Using a binomial distribution where n ⫽ 10, p ⫽ 0.337, we find P共0兲 ⫽ 0.0164. It is not impossible to obtain zero clinical pregnancies out of 10 ART cycles. However, it is rather unlikely to occur. 3. (a) Using n ⫽ 10, p ⫽ 0.230, P共8兲 ⫽ 0.0002 Suspicious, because the probability is very small. (b) Using n ⫽ 10, p ⫽ 0.192, P共0兲 ⫽ 0.1186 Not suspicious, because the probability is not that small.
CHAPTER 5
REAL STATISTICS–REAL DECISIONS 1. (a) z ⫽
x ⫺ 11.55 ⫺ 11.56 ⫽ ⫽ ⫺0.2 0.05
P共not detecting shift兲 ⫽ P共x < 11.55兲 ⫽ P共z < ⫺0.2兲 ⫽ 0.4207 (b) n ⫽ 15
p ⫽ 0.4207
q ⫽ 0.5793
np ⫽ 共15兲共0.4207兲 ⬇ 6.3105 > 5
and
nq ⫽ 共15兲共0.5793兲 ⬇ 8.6895 > 5
⫽ 冪npq ⫽ 冪共15兲共0.4207兲共0.5793兲 ⫽ 冪3.656 ⬇ 1.9121 ⫽ 6.3105 x ⫽ 0.5 ← correction for continuity z⫽
0.5 ⫺ 6.3105 ⫺5.8105 ⬇ ⬇ ⫺3.04 1.9121 1.9121
P共x ⱖ 0.5兲 ⫽ P共z ⱖ ⫺3.04兲 ⬇ 1 ⫺ 0.0012 ⫽ 0.9988 2. (a) z ⫽
x ⫺ 11.55 ⫺ 11.56 ⫺0.01 ⫽ ⫽ ⫽ ⫺0.45 0.05 0.0224 冪n 冪5
P共not detecting shift兲 ⫽ P共x < 11.55兲 ⫽ P共z < ⫺0.45兲 ⫽ 0.3264
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(b) n ⫽ 3
p ⫽ 0.3264
q ⫽ 0.6736
np ⫽ 共3兲共0.3264兲 ⬇ 0.9792 < 5
and
nq ⫽ 共3兲共0.6736兲 ⬇ 2.0208 < 5
P共at least 1兲 ⫽ P共1兲 ⫹ P共2兲 ⫹ P共3兲 ⫽ 0.4443 ⫹ 0.2153 ⫹ 0.0348 ⫽ 0.6944 (c) The mean is more sensitive to change. 3. Answers will vary.
CHAPTER 6
REAL STATISTICS–REAL DECISIONS 1. (a) No, there has not been a change in the mean concentration levels because the confidence interval for year 1 overlaps with the confidence interval for year 2. (b) Yes, there has been a change in the mean concentration levels because the confidence interval for year 2 does not overlap with the confidence interval for year 3. (c) Yes, there has been a change in the mean concentration levels because the confidence interval for year 1 does not overlap with the confidence interval for year 3. 2. Due to the fact the CI from year 1 does not overlap the CI from year 3, it is very likely that the efforts to reduce the concentration of formaldehyde in the air are significant over the 3year period. 3. (a) Used the sampling distribution of the sample means due to the fact that the “mean concentration” was used. The point estimate is the most unbiased estimate of the population mean. (b) No, it is more likely the sample standard deviation of the current year’s sample was used. Typically is unknown.
CHAPTER 7
REAL STATISTICS–REAL DECISIONS 1. (a) Take a random sample to get a diverse group of people. Stratified random sampling (b) Random sample (c) Answers will vary.
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REAL STATISTICS – REAL DECISIONS
2. H0 : p ⱖ 0.40; Ha : p < 0.40 (claim) z0 ⫽ ⫺1.282 p⫽ ^
x 15 ⫽ ⬇ 0.469 n 32 p⫺p
0.469 ⫺ 0.40
^
z⫽
⫽
冪pzn 冪共0.4032兲共0.60兲
⫽
0.069 ⫽ 0.794 0.087
Fail to reject H0 . There is not enough evidence to support the claim. 3. H0 : ⱖ 60 (claim); Ha : < 60 x ⫽ 53.067 s ⫽ 19.830 n ⫽ 15 t0 ⫽ ⫺1.345 ⫺ 53.067 ⫺ 60 ⫺6.933 t⫽x ⫽ ⫽ ⫽ ⫺1.354 s 19.83 5.120 冪n 冪15 Reject H0 . There is enough evidence to reject the claim. 4. There is not enough evidence to conclude the proportions of Americans who think Social Security will have money to provide their benefits is less than 40%. There is enough evidence to conclude the mean age of those individuals responding “yes” is less than 60. (Answers will vary.)
CHAPTER 8
REAL STATISTICS–– REAL DECISIONS 1. (a) Take a simple random sample of records today and 10 years ago from a random sample of hospitals. Cluster sampling (b) Answers will vary. (c) Answers will vary. 2. ttest; Independent 3. H0: 1 ⫽ 2 ; Ha: 1 ⫽ 2 (claim) d.f. ⫽ min{n1 ⫺ 1, n2 ⫺ 1} ⫽ 19 t0 ⫽ ± 2.093 t⫽
共x1 ⫺ x2 兲 ⫺ 共 1 ⫺ 2 兲
冪n
s 21 1
⫹
s 22 n2
⫽
共3.7 ⫺ 3.2兲 ⫺ 共0兲
冪
共1.5兲 共0.8兲 ⫹ 20 25 2
2
⫽
0.5 冪0.1381
⬇ 1.345
Fail to reject H0 . There is not enough evidence to support the claim.
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REAL STATISTICS – REAL DECISIONS
CHAPTER 9
REAL STATISTICS–– REAL DECISIONS y
1. (a)
Negative linear correlation
Mean crawling age (in weeks)
34 33 32 31 30 29 28 x 30 40 50 60 70 80
Mean monthly temperature (in °F)
(b)
Mean monthly temperature, x
Mean crawling age, y
xy
x2
y2
33 30 33 37 48 57 66 73 72 63 52 39
33.64 32.82 33.83 33.35 33.38 32.32 29.84 30.52 29.70 31.84 28.58 31.44
1110.12 984.60 1116.39 1233.95 1602.24 1842.24 1969.44 2227.96 2138.40 2005.92 1486.16 1226.16
1089.00 900.00 1089.00 1369.00 2304.00 3249.00 4356.00 5329.00 5184.00 3969.00 2704.00 1521.00
1131.65 1077.15 1144.47 1112.22 1114.22 1044.58 890.43 931.47 882.09 1013.79 816.82 988.47
兺x ⫽ 603
兺y ⫽ 381.26
兺xy ⫽ 18,943.58
兺x 2 ⫽ 33,063
兺y2 ⫽ 12,147.36
r⫽
n共兺xy兲 ⫺ 共兺x兲共兺y兲 冪n兺x2 ⫺ 共兺x兲2 冪n兺y2 ⫺ 共兺y兲2
⫽
12共18,943.58兲 ⫺ 共603兲共381.26兲 冪12共33,063兲 ⫺ 共603兲2 冪12共12,147.36兲 ⫺ 共381.26兲2
⫽ ⫺0.700
(c) H0: ⫽ 0; Ha: ⫽ 0 n ⫽ 12 ⇒ cv ⫽ 0.576
ⱍrⱍ ⫽ 0.700 > 0.576 ⇒ Reject H0. There is a significant linear correlation. (d) m ⫽
n共兺xy兲 ⫺ 共兺x兲共兺y兲 12共18,943.58兲 ⫺ 共603兲共381.26兲 ⫽ ⫽ ⫺0.078 n共兺x2兲 ⫺ 共兺x兲2 12共33,063兲 ⫺ 共603兲2
b ⫽ y ⫺ mx ⫽
603 ⫺ 共⫺0.078兲冢 ⫽ 35.678 冢381.26 12 冣 12 冣
y ⫽ ⫺0.078x ⫹ 35.678 ^
Mean crawling age (in weeks)
y 34 33 32 31 30 29 28 x 30 40 50 60 70 80
Mean monthly temperature (in °F)
The regression line appears to be a good fit. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

REAL STATISTICS – REAL DECISIONS
(e) Because the regression line is a good fit, it is reasonable to use temperature to predict mean crawling age. (f) r2 ⫽ 0.490 se ⫽ 1.319 49% of the variation in y is explained by the regression line. The standard error of estimate of mean crawling age for a specific mean monthly temperature is 1.319 weeks. 2. Agree. Due to high correlation, parents can “consider” the temperature when determining when their baby will crawl. However, the parents should not “conclude” that temperature is the other variable related to mean crawling age.
CHAPTER 10
REAL STATISTICS–– REAL DECISIONS 1. Ages Under 20 20–29 30–39 40–49 50–59 60–69 70+
Distribution
Observed
Expected
共O ⴚ E兲2 E
1% 13% 16% 19% 16% 13% 22%
30 200 300 270 150 40 10
10 130 160 190 160 130 220
40.000 37.692 122.500 33.684 0.625 62.308 200.455
2 ⫽ 497.26
H0: Distribution of ages is as shown in the table above. (claim) Ha: Distribution of ages differs from the claimed distribution. d.f. ⫽ n ⫺ 1 ⫽ 6 20 ⫽ 16.812 Reject H0. There is enough evidence at the 1% level to conclude the distribution of ages of telemarketing fraud victims differs from the survey. 2. (a) Type of Fraud Sweepstakes Credit cards Total
Age Under 20
20 –29
30–39
40–49
50–59
60–69
70–79
80+
Total
10 (15)
60 (120)
70 (165)
130 (185)
90 (135)
160 (115)
280 (155)
200 (110)
1000
20 (15)
180 (120)
260 (165)
240 (185)
180 (135)
70 (115)
30 (155)
20 (110)
1000
30
240
330
370
270
230
310
220
2000
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REAL STATISTICS – REAL DECISIONS
(b) H0: The type of fraud is independent of the victim’s age. Ha : The type of fraud is dependent on the victim’s age. d.f. ⫽ 共r ⫺ 1兲共c ⫺ 1兲 ⫽ 共1兲共7兲 ⫽ 7
20 ⫽ 18.475
共O ⫺ E 兲 2 ⬇ 619.533 E Reject H0 . There is enough evidence at the 1% significance level to conclude that the victim’s ages and type of fraud are dependent. 2 ⫽ 兺
CHAPTER 11
REAL STATISTICS–– REAL DECISIONS 1. (a) random sample (b) Answers will vary. (c) Answers will vary. 2. (a) Answers will vary. For example, ask “Is the data a random sample?” and “Is the data normally distributed?” (b) Sign test; You need to use the nonparametric test because nothing is known about the shape of the population. (c) H0 : median ⱖ 4.0; H0 : median < 4.0 (claim) (d) ␣ ⫽ 0.05 n ⫽ 19 cv ⫽ 5 x⫽7 Fail to reject H0 . There is not enough evidence at the 5% significance level to support the claim that the median tenure is less than 4.0 years. 3. (a) Because the data from each sample appears nonnormal, the Wilcoxon Rank Sum test should be used. (b) H0 : The median tenure for male workers is less than or equal to the median tenure for female workers. Ha : The median tenure for male workers is greater than the median tenure for female workers. (claim)
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REAL STATISTICS – REAL DECISIONS
(c) R ⫽ 164.5
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 12共12 ⫹ 14 ⫹ 1兲 ⫽ ⫽ 162 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪共12兲共14兲共1212⫹ 14 ⫹ 1兲 ⬇ 19.442
R ⫺ R 164.5 ⫺ 162 ⫽ ⬇ 0.129 R 19.442
Using ␣ ⫽ 0.05, z0 ⫽ ± 1.645 Fail to reject H0 . There is not enough evidence at the 5% significance level to conclude that the median tenure for male workers is greater than the median tenure for female workers.
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379

TECHNOLOGY
CHAPTER 1
TECHNOLOGY: USING TECHNOLOGY IN STATISTICS 1. From a list of numbers ranging from 1 to 86, randomly select eight of them. Answers will vary. 2. From a list of numbers ranging from 1 to 300, randomly select 25 of them. Answers will vary. 3. From a list of numbers ranging from 0 to 9, randomly select five of them. Repeat this two times. Average these three numbers and compare with the population average of 4.5. Answers will vary. 4. The average of the numbers 0 to 40 is 20. From a list of numbers ranging from 0 to 40, randomly select seven of them. Repeat this three times. Average these three numbers and compare with the population average of 20. Answers will vary. 5. Answers will vary. 6. No, we would anticipate 101’s, 102’s, 103’s, 104’s, 105’s, and 106’s. An inference that we might draw from the results is that the die is not a fair die. 7. Answers will vary. 8. No, we would anticipate 50 heads and 50 tails. An inference that we might draw from the results is that the coin is not fair. 9. The analyst could survey 10 counties by assigning each county a number from 1 to 47 and using a random number generator to find 10 numbers that correspond to certain counties.
CHAPTER 2
TECHNOLOGY: MONTHLY MILK PRODUCTION 1. x ⫽ 2270.5 2. s ⫽ 653.2 Lower limit
Upper limit
Frequency
1147 1647 2147 2647 3147 3647 4147
1646 2146 2646 3146 3646 4146 4646
7 15 13 11 3 0 1
4.
4396.5
3896.5
1396.5 1896.5
Frequency
Monthly Milk Production 16 14 12 10 8 6 4 2 2396.5 2896.5 3396.5
3.
Monthly milk production (in pounds)
The distribution does not appear to be bellshaped. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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TECHNOLOGY
5. 74% of the entries were within one standard deviation of the mean 共1617.3, 2923.7兲. 98% of the entries were within two standard deviations of the mean 共964.1, 3576.9兲. 6. x ⫽ 2316.5 7. s ⬇ 641.75 8. Answers will vary.
CHAPTER 3
TECHNOLOGY: SIMULATION: COMPOSING MOZART VARIATIONS WITH DICE 1. 2 ⫹ 11 ⫽ 13 phrases were written. 2. 共11兲7 3. (a)
⭈ 2 ⭈ 共11兲7 ⭈ 2 ⫽ 1.518999334333 ⫻ 1015
1 ⬇ 0.091 11
(b) Results will vary. 4. (a) P共option 6, 7, or 8 for 1st bar兲 ⫽
3 ⫽ 0.273 11
P共option 6, 7, or 8 for all 14 bars兲 ⫽
冢113 冣
14
⬇ 0.000000012595
(b) Results will vary. 5. (a) P共1兲 ⫽
1 36
P共2兲 ⫽
2 36
P共3兲 ⫽
3 36
P共4兲 ⫽
4 36
P共5兲 ⫽
5 36
P共6兲 ⫽
6 36
P共7兲 ⫽
5 36
P共8兲 ⫽
4 36
P共9兲 ⫽
3 36
P共10兲 ⫽
2 36
P共11兲 ⫽
1 36
(b) Results will vary. 6. (a) P共option 6, 7, or 8 for 1st bar兲 ⫽
15 ⬇ 0.417 36
P共option 6, 7, or 8 for all 14 bars兲 ⫽
冢 冣 15 36
14
⬇ 0.00000475
(b) Results will vary.
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TECHNOLOGY
CHAPTER 4
TECHNOLOGY: USING POISSON DISTRIBUTIONS AS QUEUING MODELS 1.
x
P冇 x)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0.0183 0.0733 0.1465 0.1954 0.1954 0.1563 0.1042 0.0595 0.0298 0.0132 0.0053 0.0019 0.0006 0.0002 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
2. (a) (See part b) 1, 2, 4, 7 (b) Minute
Customers Entering Store
Total Customers in Store
Customers Serviced
Customers Remaining
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
3 3 3 3 5 5 6 7 3 6 3 5 6 3 4 6 2 2 4 1
3 3 3 3 5 6 8 11 10 12 11 12 14 13 13 15 13 11 11 8
3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
0 0 0 0 1 2 4 7 6 8 7 8 10 9 9 11 9 7 7 4
3. Answers will vary. 4. 20 customers (an additional 1 customer would be forced to wait in line/minute) 5. Answers will vary. 6. P共10兲 ⫽ 0.0181 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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TECHNOLOGY
7. (a) It makes no difference if the customers were arriving during the 1st minute or the 3rd minute. The mean number of arrivals will still be 4 customers/minute. P共3 ⱕ x ⱕ 5兲 ⫽ P共3兲 ⫹ P共4兲 ⫹ P共5兲 ⬇ 0.1954 ⫹ 0.1954 ⫹ 0.1563 ⫽ 0.5471 (b) P共x > 4兲 ⫽ 1 ⫺ P共x ⱕ 4兲 ⫽ 1 ⫺ 关P共0兲 ⫹ P共1兲 ⫹ . . . ⫹ P共4兲兴 ⫽ 1 ⫺ 关0.6289兴 ⫽ 0.3711 (c) P共x > 4 during each of the first four minutes兲 ⫽ 共0.3711兲4 ⬇ 0.019
CHAPTER 5
TECHNOLOGY: AGE DISTRIBUTION IN THE UNITED STATES 1. ⬇ 36.59 2. x of the 36 sample means ⬇ 36.209 This agrees with the Central Limit Theorem. 3. No, the distribution of ages appears to be positively skewed. 4. 0.30 0.25 0.20 0.15 0.10 0.05 29.06 30.91 32.76 34.61 36.46 38.31 40.16 42.01 43.86
Relative frequency
Distribution of 36 Sample Means
Mean age (in years)
The distribution is approximately bellshaped and symmetrical. This agrees with the Central Limit Theorem. 5. ⬇ 22.499 6. of the 36 sample means ⫽ 3.552.
22.499 ⫽ ⬇ 3.750 冪n 冪36 This agrees with the Central Limit Theorem.
CHAPTER 6
TECHNOLOGY: MOST ADMIRED POLLS 1. p ⫽ 0.13 ^
冪pnq ⫽ 0.013 ± 1.96冪0.0131010⭈ 0.87 ⫽ 0.13 ± 0.021 ⬇ 共0.109, 0.151兲 ^ ^
p ± zc ^
2. No, the proportion is 13% plus or minus 2.1%.
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TECHNOLOGY
3. p ⫽ 0.09 ^
冪
p ± zc ^
⭈ 0.91 ⫽ 0.09 ± 0.018 ⬇ 共0.072, 0.108兲 冪0.091010
pq ⫽ 0.09 ± 1.96 n ^ ^
4. Answers will vary. 5. No, because the 95% CI does not contain 10%.
CHAPTER 7
TECHNOLOGY: THE CASE OF THE VANISHING WOMEN 1. Reject H0 with a Pvalue less than 0.01. 2. Type I error 3. Sampling process was nonrandom. 4. (a) H0: p ⫽ 0.2914 (claim); Ha: p ⫽ 0.2914 (b) Test and Confidence Interval for One Proportion Test of p ⫽ 0.2914 vs p not ⫽ 0.2914 Exact Sample 1
X 9
N 100
Sample p 0.090000
99.0% CI 共0.016284, 0.163716兲
ZValue ⫺4.43
PValue 0.000
(c) Reject H0. (d) It was highly unlikely that random selection produced a sample of size 100 that contained only 9 women.
CHAPTER 8
TECHNOLOGY: TAILS OVER HEADS 1. Test and Confidence Interval for One Proportion Test of p ⫽ 0.5 vs. p not ⫽ 0.5 Exact Sample 1
X 5772
N 11902
Sample p 0.484961
95.0% CI 共0.47598, 0.49394兲
PValue 0.001
Reject H0: p ⫽ 0.5 2. Yes, obtaining 5772 heads is a very uncommon occurrence (see sampling distribution). The coins might not be fair. 3.
1 500
⭈ 100 ⫽ 0.2%
4. z ⫽ 8.802 → Fail to reject H0: 1 ⫽ 2 There is sufficient evidence at the 5% significance level to suppport the claim that there is a difference in the mint dates in Philadelphia and Denver. (Answers will vary.)
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TECHNOLOGY
5. z ⫽ 1.010 → Fail to reject H0: 1 ⫽ 2 There is insufficient evidence at the 5% significance level to support the claim that there is a difference in the value of the coins in Philadelphia and Denver. (Answers will vary.)
CHAPTER 9
TECHNOLOGY: NUTRIENTS IN BREAKFAST CEREALS y 2.0
Fat (in grams)
16 12 8 4
1.5 1.0 0.5
x
x
40 35 30 25 20 x
100 120 140 160 180
100 120 140 160 180
Calories
Calories
Calories
y
2.0 1.5 1.0 0.5 x 5
10
15
Carbohydrates (in grams)
y
Fat (in grams)
45
100 120 140 160 180
y
45 40 35 30 25 20 x
20
5
Sugar (in grams)
10
15
20
Sugar (in grams)
Carbohydrates (in grams)
Sugar (in grams)
20
y
Carbohydrates (in grams)
y
1.
45 40 35 30 25 20 x 0.5
1.0
1.5
2.0
Fat (in grams)
2. (calories, sugar), (calories, carbohydrates), (sugar, carbohydrates) 3. (calories, sugar): (calories, fat): (calories, carbohydrates): (sugar, fat): (sugar, carbohydrates): (fat, carbohydrates):
r ⬇ 0.566 r ⬇ 0.194 r ⬇ 0.927 r ⬇ 0.311 r ⬇ 0.544 r ⬇ ⫺0.021
Largest r: (calories, carbohydrates) 4. (a) y ⫽ ⫺8.823 ⫹ 0.151x ^
(b) y ⫽ ⫺3.417 ⫹ 0.252x ^
5. (a) y ⫽ ⫺8.823 ⫹ 0.151共120兲 ⫽ 9.297 grams ^
(b) y ⫽ ⫺3.417 ⫹ 0.252共120兲 ⫽ 26.823 grams ^
6. (a) C ⫽ 22.055 ⫺ 0.043S ⫹ 6.644F ⫹ 3.455R (b) C ⫽ 29.427 ⫹ 0.329S ⫹ 3.240R 7. C ⫽ 29.427 ⫹ 0.329共10兲 ⫹ 3.240共25兲 ⫽ 113.717 calories
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TECHNOLOGY
CHAPTER 10
TECHNOLOGY: TEACHER SALARIES 1. Yes, since the salaries are from different states, it is safe to assume that the samples are independent. 2. Using a technology tool, it appears that all three samples were taken from approximately normal populations. 3. H0: 21 ⫽ 22 ; Ha: 21 ⫽ 22 共F0 ⫽ 2.86兲
共CA, OH兲 → F ⬇ 1.048 → Fail to reject H0. 共CA, WY兲 → F ⬇ 1.114 → Fail to reject H0. 共OH, WY兲 → F ⬇ 1.167 → Fail to reject H0. 4. The three conditions for a oneway ANOVA test are satisfied. H0: 1 ⫽ 2 ⫽ 3 (claim) Ha: At least one mean is different from the others. Variation
Sum of Squares
Degrees of Freedom
Between Within
2,345,804,805 2,738,312,005
2 45
Mean Square
F
1,172,902,402 19,275 60,851,377.9
F ⬇ 19.275 Reject H0. There is not enough evidence at the 5% significance level to reject the claim that the mean salaries are the same. 5. The samples are independent. H0: 21 ⫽ 22 ; Ha: 21 ⫽ 22 共F0 ⫽ 2.86兲
共AK, NV兲 → F ⬇ 1.223 → Fail to reject H0. 共AK, NY兲 → F ⬇ 3.891 → Reject H0. 共NV, NY兲 → F ⬇ 4.760 → Reject H0. The sample from the New York shows some signs of being drawn from a nonnormal population. The three conditions for oneway ANOVA are not satisfied.
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TECHNOLOGY
CHAPTER 11
TECHNOLOGY: U.S. INCOME AND ECONOMIC RESEARCH 1.
Annual Income of People (in dollars) West South Midwest Northeast
10,000
20,000
30,000
40,000
50,000
Income (in dollars)
The median annual incomes do not appear to differ between regions. 2. H0: median ⱕ $25,000 Ha: median > $25,000 (claim) The critical value is 2. x⫽1 Reject H0. There is enough evidence to support the claim. 3. H0: There is no difference in incomes in the Northeast and South. (claim) Ha: There is a difference in incomes in the Northeast and South. The critical values are z0 ⫽ ± 1.96. R ⫽ 164.5
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 12共12 ⫹ 12 ⫹ 1兲 ⫽ ⫽ 150 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪共12兲共12兲共1212⫹ 12 ⫹ 1兲 ⬇ 17.3
R ⫺ R 164.5 ⫺ 150 ⬇ 0.838 ⫽ R 17.3
Fail to reject H0. There is not enough evidence to reject the claim. 4. H0: There is no difference in the incomes for all four regions. (claim) Ha: There is a difference in the incomes for all four regions. H ⫽ 1.03 Fail to reject H0. There is not enough evidence to reject the claim.
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TECHNOLOGY
5. H0: There is no difference in the incomes for all four regions. (claim) Ha: There is a difference in the incomes for all four regions. Analysis of Variance Source
Sum of Squares
Degrees of Freedom
Mean Square
Factor Error
95,270,310 1.903 ⫻ 10 9
3 44
31,756,770 43,247,860
Total
1.998 ⫻ 10 9
47
F
P
0.73
0.537
F ⬇ 0.74 → Pvalue ⫽ 0.537 Fail to reject H0. There is not enough evidence to reject the claim. 6. (Boxandwhisker plot) Annual Income of Families (in dollars) West South Midwest Northeast
20,000
40,000
60,000
80,000 100,000 120,000
Income (in dollars)
The median family incomes appear to be higher in the Northeast and lower in the South. (Wilcoxon rank sum test) H0: There is no difference in incomes in the Northeast and South. (claim) Ha: There is a difference in incomes in the Northeast and South. The critical values are z0 ⫽ ± 1.96. R ⫽ 281
R ⫽
n1共n1 ⫹ n2 ⫹ 1兲 15共15 ⫹ 15 ⫹ 1兲 ⫽ ⫽ 232.5 2 2
R ⫽
冪n n 共n 12⫹ n
z⫽
1 2
1
2
⫹ 1兲
⫽
冪共15兲共15兲共1512⫹ 15 ⫹ 1兲 ⬇ 24.1
R ⫺ R 281.0 ⫺ 232.5 ⫽ ⬇ 2.012 R 24.1
Reject H0. There is enough evidence to reject the claim. (KruskalWallis test) H0: There is no difference in incomes for all four regions. (claim) Ha: There is a difference in incomes for all four regions. H ⫽ 6.09 Fail to reject H0. There is not enough evidence to reject the claim.
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TECHNOLOGY
(ANOVA test) H0: There is no difference in incomes for all four regions. (claim) Ha: There is a difference in incomes for all four regions. Analysis of Variance Degrees of Freedom
Mean Square
Factor Error
10 9
1.283 ⫻ 1.172 ⫻ 10 10
3 56
427,704,638 209,309,693
Total
1.300 ⫻ 10 10
59
Source
Sum of Squares
F
P
2.04
0.118
F ⬇ 1.93 → pvalue ⫽ 0.135 Fail to reject H0. There is not enough evidence to reject the claim.
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